Artin’s Algebra

Matrices

Basic Operations

Let \(m\) and \(n\) be positive integers. An \(m \times n\) matrix is a collection of \(mn\) numbers arranged in a rectangular array.

::: {.example #unnamed-chunk-1} \[A:=\begin{bmatrix} 8 & 0 & 3\\ 78 & -5 & 2 \end{bmatrix}\] \(A\) is \(2 \times 3\) matrix.(two rows and three columns)

::: The numbers in a matrix are the matrix entries. They may be denoted by \(a_{ij}\), where \(i\) and \(j\) are indices (integers) with \(1 < i < m\) and \(1 < j < n\). The index \(i\) represents the row index, and \(j\) represents the column index. So \(a_{ij}\) is the entry that appears in the \(i\)th row and \(j\)th column of the matrix.

\(~\)

Group Theory

Laws of Compositions

A law o f composition on a set \(S\) is any rule for combining pairs \(a, b\) of elements of \(S\) to get another element, say \(p\), of \(S\).

• Some models for this concept are addition and multiplication of real numbers.
• Matrix multiplication on the set of \(n \times n\) matrices is another example.

Formally, a law of composition is a function of two variables, or a map, \[S\times S \to S\] Here, \(S \times S\) denotes, as always, the product set, whose elements are pairs \(a, b\) of elements of \(S\).

The element obtained by applying the law to a pair \(a, b\) is usually written using a notation resembling one used for multiplication or addition: \[p = ab, a \times b, a \circ b, a + b\], or whatever, a choice being made for the particular law in question. The element \(p\) may be called the product or the sum of a and b, depending on the notation chosen.

Groups and Subgroups

A group is a set \(G\) together with a law of composition that has the following properties:

• The law of composition is associative: \((ab)c = a(bc)\) for all \(a\), \(b\), \(c\) in \(G\).
• \(G\) contains an identity element \(1\), such that \(la = a\) and \(al = a\) for all \(a\) in \(G\).
• Every element \(a\) of \(G\) has an inverse, an element \(b\) such that \(ab = 1\) and \(ba = 1\).

Notation: If set \(G\) with composition \(\cdot\) is a group, then we denote it by \((G,\cdot)\) ::: {.definition #unnamed-chunk-5 name=“Ableian Group”} Group G is called abliean if its law of composition is commutative. i.e. \[\forall x\in G, xy=yx\] :::

• \(\mathbb{R}^\times:=\mathbb{R}\setminus \{0\}\) is an abliean group under multiplication
• \(GL_n(\mathbb{R}):=\{A\in M_n(\mathbb{R}): \text{ A is invertible}\}\) with materix multiplication is non-abliean group. This group is called general linear group.

The order of a group \(G\) is the number of elements that it contains. We \[| G | := \text{number of elements of }G=\text{ the order of G}\]

If the order is finite, \(G\) is said to be a finite group. If not, \(G\) is an infinite group.

Here is our notation for some familiar infinite abelian groups:

• \((\mathbb{Z},+)\) :The set of integers, with addition as itslaw of composition (the additive group of integers)
• \((\mathbb{R},+)\) :The set of real numbers, with addition as its law of composition (the additive group of real numbers)
• \((\mathbb{R}^\times,\times)\) :The set of nonzero real numbers, with multiplication as its law of composition(the multiplicative group) \((\mathbb{C},+)\) :the set of complex numbers, with addition as its law of composition (the additive group of complex numbers)
• \((\mathbb{C}^\times,\times)\) :The set of nonzero complex numbers, with multiplication as its law of composition(the multiplicative group of complex numbers)

Let \(G\) be group and let \(a, b, c \in G\) whose law of composition is written multiplicatively.

• If \(ab = ac\) or if \(ba = ca\), then \(b = c\).
• If \(a b = a\) or if \(ba = a\), then \(b = 1\)

Multiply both sides of \(ab=ac\) on the left by\(a^{-1}\) to obtain \(b = c\). The other proofs are analogous.

As you saw \(a^{-1}\) plays amajor rule in above proof. So the cancelation rule does not holds when elemenet \(a\) is not invertible.

\[\begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 0\end{bmatrix}= \begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & 1\end{bmatrix}\]

Let \(T\) be a set and \(G:=\{f:T\to T: f \text{ is a bijection}\}\). Then \(G\) with composition is a group. We use notation $sys(T) $ to denote the

The group of permutations of the set of indices \(\{l, 2, ... , n\}\) is called the symmetric group,and is denoted by \(S_n\). Then \(|S_n|=n !\). So, \(S_n\) is a finite group of order \(n!\).

Let’s disscuss some individualcases for n.

• \(n=2\)
  The permutations of a set \(\{1, 2\}\) of two elements are the identity \(i\) and the transposition \(\tau=(12)\). \[S_2:=\{id,(12)\}\]

\(\circ\)	\(id\)	\((12)\)
\(id\)	\(id\)	\((12)\)
\((12)\)	\((12)\)	\(id\)
——	——	—–

• \(n=3\)
  \(S_3\) has order \(3!=6\). \(S_3\) serves as a convenient example because it is the smallest group whose law of composition isn’t commutative. We will refer to it often. To describe it, we pick two particular permutations in terms of which we can write all others. We take the cyclic permutation \((123)\), and the transposition \((12)\), and label them as \(x\) and \(y\), respectively.Then \[\begin{eqnarray} x^3 &=& (123)^3=(123)(123)(123)=(123)(132)=id\\ y^2 &=& (12)^2=(12)(12)=id\\ yx &=& (12)(123)=(13)=(132)(12)=((123)(123))(12)=(123)^2(12) \end{eqnarray}\] As a summary, \[x^3=1,Y^2=1,yx=x^2y\]

Subgroups of the Additive Group of Integers

We review some elementary number theory here, in terms of subgroups of the additive group \(\mathbb{Z}^+\) of integers. To begin, we list the axioms for a subgroup when additive notation is used in the group: A subset \(S\) of a group \(G\) with law of composition written additively is a subgroup if it has these properties:

• Closure : If \(a\) and \(b\) are in \(S\), then \(a + b\) is in \(S\).
• Identity : \(0\) is in \(S\).
• Inverses : If \(a\) is in \(S\) then \(-a\) is in \(S\).

Let \(a\) be an integer different from \(O\). We denote the subset of \(\mathbb{Z}\) that consists of all multiples of \(a\) by \(a\mathbb{Z}\): \[a\mathbb{Z} := \{n \in \mathbb{ Z} | n = ka \text{ for some } k\in \mathbb{Z}\} \].

This is a subgroup of \(\mathbb{Z}^+\). Its elements can also be described as the integers divisible by \(a\).

Let \(S\) be a subgroup of the additive group \(\mathbb{Z}^+\). Either \(S\) is the trivial subgroup \(\{0\}\), or else it has the form \(a\mathbb{Z}\), where \(a\) is the smallest positive integer in \(S\).

Let \(S\) be a subgroup of \(\mathbb{Z}^+\). Then \(0\in S\), and if \(0\) is the only element of \(S\) then \(S\) is the trivial subgroup. ie.: \(S=\{0\}\)So that case is settled.

Otherwise, \(S\) contains an integer \(n\) different from \(0\), and either \(n\) or \(-n\) is positive. The third property of a subgroup tells us that \(-n\) is in \(S\), so in either case, \(S\) contains a positive integer. We must show that \(S\) is equal to \(a\mathbb{Z}\), when \(a\) is the smallest positive integer in \(S\).

We first show that \(a\mathbb{Z}\) is a subset of \(S\), in other words, that \(ka\) is in \(S\) for every integer \(k\). If \(k\) is a positive integer, then \(ka = a + a + \cdots + a\) (\(k\) terms). Since \(a\) is in \(S\), closure and induction show that \(ka\) is in \(S\). Since inverses are in \(S\), \(-ka\) is in \(S\). Finally, \(0 = 0a\) is in \(S\).

Next we show that \(S\) is a subset of \(Za\), that is, every element \(n\) of \(S\) is an integer multiple of \(a\). We use division with remainder to write \(n = qa + r\), where \(q\) and \(r\) are integers and where the remainder \(r\) is in the range \(0 < r < a\). Since \(Za\) is contained in \(S\), \(qa\) is in \(S\), and of course \(n\) is in \(S\). Since \(S\) is a subgroup, \(r = n - qa\) is in \(S\) too. Now by our choice, \(a\) is the smallest positive integer in \(S\), while the remainder \(r\) is in the range \(0 < r < a\). The only remainder that can be in \(S\) is \(0\). So \(r = 0\) and \(n\) is the integer multiple \(qa\) of \(a\).

Cyclic Groups

Homomorphisms

Let \((G, *)\) and \((G', \odot)\) be groups. A homomorphism \(\phi: G \rightarrow G'\) is a map from \(G\) to \(G'\) such that for all \(a\) and \(b\) in \(G\):

\[\begin{equation} \phi(a * b) = \phi(a) \odot \phi(b). \end{equation}\]

Examples of Homomorphisms

1. Determinant Function: \(\text{det}: GL_n(\mathbb{R}) \to \mathbb{R}^\times\)
2. Exponential Map: \(\text{exp}: (\mathbb{R}, +) \to (\mathbb{R}^\times, \times)\) defined by \(x \mapsto e^x\)
3. Map \(\phi\): \((\mathbb{Z}, +) \to G\) defined by \(\phi(n) = a^n\), where \(a\) is a given element of \(G\)
4. Absolute Value Map: \(|\cdot| : (\mathbb{C}^\times, \times) \to (\mathbb{R}^\times, \times)\)

Trivial Homomorphism

The trivial homomorphism \(\phi: G \rightarrow G'\) between any two groups maps every element of \(G\) to the identity in \(G'\).

Inclusion Map

If \(H\) is a subgroup of \(G\), the inclusion map \(i: H \rightarrow G\) defined by \(i(x) = x\) for \(x\) in \(H\) is a homomorphism.

Let \(\phi: G \rightarrow G'\) be a group homomorphism.

1. If \(a_i, \ldots, a_k\) are elements of \(G\), then \(\phi(a_i \ldots a_k) = \phi(a_i) \ldots \phi(a_k)\).

2. \(\phi\) maps the identity to the identity: \(\phi(1_G) = 1_{G'}\).

3. \(\phi\) maps inverses to inverses: \(\phi(a^{-1}) = \phi(a)^{-1}\).

Let \(\phi : G \rightarrow G'\) be a homomorphism of groups, and let \(a,b\in G\). Let \(K\ker(\phi)\). Then following conditions are equivalent:

• \(\phi(a) = \phi(b)\),
  
• \(a^{-1}b\) is in \(K\),
• \(b\) is in the coset \(aK\),
• The cosets \(bK\) and \(aK\) are equal.

• \((1)\implies (2)\).\ Suppose \(\phi(a) = \phi(b)\). Now consider \[\phi(a^{-1}b)=\phi(a^{-1})\phi(b)=\phi(a^{-1} )\phi(b)=1\] Thus, \(a^{-1}b\in K\).
• \((2)\implies (1)\)\ If \(a^{-1}b \in K\) \[1=\phi(a^{-1}b)=\phi(a^{-1})\phi(b)=\phi(a^{-1} )\phi(b)\]. Thus, \(\phi(a)=\phi(b)\). \((3) \implies (4)\)
• Suppose \(b\) is in the coset \(aK\). Then \(b=ak\) for some

Isomorphisms

Equivalence Relations and Partitions

Sure, here is the LaTeX version of your statement:

A partition \(n\) of a set \(S\) is a subdivision of \(S\) into nonoverlapping, nonempty subsets: \[S = \bigcup_{i} A_i\] ,where \(A_i\) are disjoint nonempty subsets of \(S\).

The two sets Even and Odd partition the set of integers.

With the usual notation, the sets \[\{1\}, \{y, xy, x^2y\}, \{x, x^2\}\] form a partition of the symmetric group \(S_3\).

An equivalence relation on a set \(S\) is a relation that holds between certain pairs of elements.

• Reflexive: For all \(a\), \(a \sim a\).
• Symmetric: If \(a \sim b\) , then \(b \sim a\).
• Transitive: If \(a \sim b\) and \(b \sim c\) , then \(a \sim c\).

Two triangles are said to be congruent if their sides have the same length and angles have same measure.
Congruence of triangles is an example of an equivalence relation on the set of triangles in the plane. If \(A , B,\) and \(C\) are triangles, and if \(A\) is congruent to \(B\) and \(B\) is congruent to \(C\), then \(A\) is congruent to \(C\). It is very easy to check three equivalnace propoties. I am not going to do this.

Conjugacy is an equivalence relation on a group. Let \(G\) be garoup Two group elements are conjugate, \(a\sim b\), if \(b = gag^{- 1}\) for some \(g\in G\).

• Reflexive: Observe that \(a=aaa^{-1}\) then, \(a\sim a\).
• Symmetric: Suppose that \(a\sim b\), then \(b=gag^{-1}\). Thus, \(a=g^{-1}bg\). Hence, \(b \sim a\).
• Transitive: Suppose that \(a\sim b\) and \(b \sim c\). This means that \(b = g_1ag_1^{-1}\) and \(c = g_2bg_2^{-1}\) for some group elements \(g_1\) and \(g_2\). Then \(c = g_2(g_1ag_1^{-1})g_2^{-1} = (g_2 g_1)a (g_2 g_1),\)  so \(a \sim c\)

An equivalence relation on a set S determines a partition of S, and conversely.

• (\(\Longrightarrow\)): Given a partition of \(S\), the corresponding equivalence relation is defined by the rule that \(a \sim b\) if \(a\) and \(b\) lie in the same subset of the partition. The axioms for an equivalence relation are obviously satisfied.
• (\(\Longrightarrow\)): Given an equivalence relation (\(\sim\)), one defines a partition this way:
  The subset that contains \(a\) is the set of all elements \(b\) such that \(a \sim b\). This subset is called the equivalence class of a. We’ll denote it by \(C_a\) here \[C_a = \{b \in S | a\sim b\}\] The next lemma completes the proof of the proposition

Given an equivalence relation on a set \(S\), the subsets of \(S\) that are equivalence classes partition \(S\).

This is an important point, so we will check it carefully. We must remember that the notation \(C_a\) stands for a subset defined in a certain way. The partition consists of the subsets, and several notations may describe the same subset.

• Non-emptyness: The reflexive axiom tells us that \(a\) is in its equivalence class. Therefore, the class \(C_a\) is nonempty.

• Union is whole set: Since \(a\) can be any element, the union of the equivalence classes is the whole set \(S\).

• Disjoint property: To show this, we prove follwoing calim:

  • Claim: \(\text{If } C_a \text{ and } C_b \text{ have an element in common, then } C_a = C_b\).
    Since we can interchange the roles of \(a\) and \(b\), it will suffice to show that if \(C_a\) and \(C_b\) have an element, say \(d\), in common, then \(C_b \subseteq C_a\).
    Sippose that \(x\) is in \(C_b\), then \(b \sim x\). Since \(d\) is in both sets, \(a \sim d\) and \(b \sim d\), and the symmetry property tells us that \(d \sim b\). So we have \(a \sim d\), \(d \sim b\), and \(b \sim x\). Two applications of transitivity show that \(a \sim x\), and therefore, \(x\) is in \(C_a\).

The relation on a group defined by \(a ~ b\) if \(a\) and \(b\) are elements of the same order is an equivalence relation. ( Trivial.)

The corresponding partition for the symmetric group S3 are \[\{1\}, \{y, xy, x^2y\}, \{x, x^2\}.\]

If a partition of a set \(S\) is given, we may construct a new set \(\overline{S}\) whose elements are the subsets. We imagine putting the subsets into separate piles, and we regard the piles as the elements of our new set \(\overline{S}\). It seems advisable to have a notation to distinguish a subset from the element of the set \(S\) (the pile) that it represents. If \(U\) is a subset, we will denote by \([U]\) the corresponding element of \(S\).

If \(S\) is the set of integers and if \(\text{Even}\) and \(\text{Odd}\) denote the subsets of even and odd integers, respectively, then \(\overline{S}\) contains the two elements \([\text{Even}]\) and \([\text{Odd}]\).

We will use this notation more generally. When we want to regard a subset \(U\) of \(S\) as an element of a set of subsets of \(S\), we denote it by \([U]\).

When an equivalence relation on \(S\) is given, the equivalence classes form a partition, and we obtain a new set \(\overline{S}\) whose elements are the equivalence classes \([C_a]\). We can think of the elements of this new set in another way, as the set obtained by changing what we mean by equality among elements. If \(a\) and \(b\) are in \(S\), we interpret \(a \sim b\) to mean that \(a\) and \(b\) become equal in \(\overline{S}\), because \(C_a = C_b\). With this way of looking at it, the difference between the two sets \(S\) and \(\overline{S}\) is that in \(\overline{S}\) more elements have been declared “equal,” i.e., equivalent. It seems to me that we often treat congruent triangles this way in school.

For any equivalence relation, there is a natural surjective map \[\begin{equation} \quad \pi : S \to \overline{S} \end{equation}\] that maps an element \(a\) of \(S\) to its equivalence class: \(\pi(a) = [C_a]\). When we want to regard \(\overline{S}\) as the set obtained from \(S\) by changing the notion of equality, it will be convenient to denote the element \([C_a]\) of \(\overline{S}\) by the symbol \(a\). Then the map \(\pi\) becomes \[\begin{equation} \pi(a) = \overline{a} \end{equation}\]

We can work in \(\overline{S}\) with the symbols used for elements of \(S\), but with bars over them to remind us of the new rule: \[\begin{equation} \text{If } a \text{ and } b \text{ are in } S, \text{ then } \overline{a} = \overline{b} \text{ means } a \sim b. \end{equation}\]

A disadvantage of this bar notation is that many symbols represent the same element of \(S\). Sometimes this disadvantage can be overcome by choosing a particular element, a representative element, in each equivalence class. For example, the even and the odd integers are often represented by \(\overline{0}\) and \(\overline{1}\): \[\begin{equation} \{[\text{Even}], [\text{Odd}]\} = \{ \overline{0}, \overline{1}\}. \end{equation}\]

Though the pile picture may be easier to grasp at first, the second way of viewing \(S\) is often better because the bar notation is easier to manipulate algebraically.

The Equivalence Relation Defined by a Map

Any map of sets \(f : S \to T\) gives us an equivalence relation on its domain \(S\). It is defined by the rule: \(a \sim b\) if \(f(a) = f(b)\).

The inverse image of an element \(t\) of \(T\) is the subset of \(S\) consisting of all elements \(s\) such that \(f(s) = t\). It is denoted symbolically as \[\begin{equation} f^{-1}(t) = \{s \in S \mid f(s) = t\}. \end{equation}\]

The inverse images are also called the fibres of the map \(f\) , and the fibres that are not empty are the equivalence classes for the relation defined above.

This is symbolic notation. Please remember that unless \(f\) is bijective, \(f^{-1}\) will not be a map.

Here the set S of equivalence classes has another incarnation, as the image of the map. The elements of the image correspond bijectively to the nonempty fibres, which are the equivalence classes

Let’s consider the absolute value map from the complex numbers \(\mathbb{C}^\times\) to the positive real numbers \(\mathbb{R}^+\):

\[ f: \mathbb{C}^\times \to \mathbb{R}^+, \quad f(z) = |z| \]

The fiber of an element \(t\) in \(\mathbb{R}^+\) is the subset of \(\mathbb{C}^\times\) consisting of all complex numbers \(z\) such that \(f(z) = |z| = t\). Symbolically, we denote the fiber as:

\[ f^{-1}(t) = \{ z \in \mathbb{C}^\times : |z| = t \} \]

This fiber represents all complex numbers with the same absolute value \(t\). Thus, fibers are circles in Complex plane.

Note that the absolute value map is surjective, so each positive real number \(t\) corresponds to a unique fiber in \(\mathbb{C}^\times\).

If \(G\) is a finite group, we can define a map \(f : G \to \mathbb{N}\) to the set \(\{1, 2, 3, \ldots\}\) of natural numbers, letting \(f(a)\) be the order of the element \(a\) in \(G\). The fibers of this map are the sets of elements with the same order (see example @ref(exm:272)).

Let \(G,G'\) be a groups and \(\phi: G \rightarrow G'\) be a group group homomorphism. The equivalence relation on \(G\) defined by \(\phi\) is usually denoted by \(\equiv\), rather than by \(\sim\), and is referred to as congruence. \[a \equiv b \text{ if }\phi(a) = \phi(b)\]

We have seen that elements \(a\) and \(b\) of \(G\) are congruent, i.e., \(\phi(a) = \phi(b)\), if and only if \(b\) is in the coset \(a K\) of the kernel \(K\) (See proposition @ref(prp:258))

Let \(K\) be the kernel of a homomorphism \(\varphi: G \to G'\). The fibre of that contains an element \(a\) of \(G\) is the coset \(a K\) of \(K\). These cosets partition the group \(G\), and they correspond to elements of the image of \(\varphi\).

The fibre of \(\varphi\) over an element \(g' \in G'\) is the set of all elements \(a \in G\) such that \(\varphi(a) = g'\). Let’s denote this fibre as \[F_{g'} = \{a \in G : \varphi(a) = g'\}\].

Now, let’s take an arbitrary element \(a \in G\) and consider the left coset \(aK = \{ak : k \in K\}\).

We know that \(K\) is the kernel of \(\varphi\), so for any \(k \in K\), we have \(\varphi(k) = 1_{G'}\) where \(1_{G'}\) is the identity element in \(G'\).

• Claim 1: \(aK=F_{\varphi(a)}\).

  • sub claim 1.1: \(aK \subseteq F_{\varphi(a)}\).
    For any \(ak \in aK\), we have \(\varphi(ak) = \varphi(a)\varphi(k) = \varphi(a)1_{G'} = \varphi(a)\).This shows that every element in the coset \(aK\) is mapped to the same element under \(\varphi\) as \(a\) itself. Therefore, the coset \(aK\) is a subset of the fibre \(F_{\varphi(a)}\).

  • sub claim 1.2 : \(F_{\varphi(a)}\subseteq aK\).
    If \(a' \in F_{\varphi(a)}\), then \(\varphi(a') = \varphi(a)\), which implies that \(\varphi((a')^{-1}a) = 1_{G'}\). This means that \((a')^{-1}a \in K\), or equivalently, \(a' \in aK\). (by proposition @ref(prp:258))Therefore, the fibre \(F_{\varphi(a)}\) is a subset of the coset \(aK\).

Since \(aK\) is a subset of \(F_{\varphi(a)}\) and \(F_{\varphi(a)}\) is a subset of \(aK\), we conclude that \(aK = F_{\varphi(a)}\). So, the fibre of \(\varphi\) that contains an element \(a\) of \(G\) is indeed the coset \(aK\) of \(K\).

• Claim 2: The cosets of a subgroup \(K\) partition the group \(G\).

A partition of a set is a collection of non-empty subsets such that every element in the set is in exactly one of these subsets.

To show that the cosets of \(K\) partition \(G\), we need to show two things:

• sub claim 2.1: Every element of \(G\) is in at least one coset of \(K\).
  Given any \(g \in G\), \(g\) is in the coset \(gK\). So, every element of \(G\) is in at least one coset of \(K\).

• sub claim 2.2: No element of \(G\) is in more than one coset of \(K\)
  Suppose \(gK\) and \(hK\) are two cosets of \(K\) and there is some element \(x \in G\) that is in both \(gK\) and \(hK\). This means that there exist \(k_1, k_2 \in K\) such that \(x = gk_1 = hk_2\) (by proposition @ref(prp:258). Then \(g = hk_2k_1^{-1}\). Thus, \(g\in gK\), which implies that \(gK = hK\). So, no element of \(G\) is in more than one coset of \(K\).

Therefore, the cosets of \(K\) partition the group \(G\).

Cosets

Modular Arithmetic

The Correspondence Theorem

Let \(\phi: G \rightarrow G^\prime\) be a group homomorphism, and let \(H\) be a subgroup of \(G\). We may restrict \(\phi\) to \(H\), obtaining a homomorphism \[\phi|_H: H \rightarrow G^\prime\]

In other words, we take the same map but restrict its domain.

Notation: We use this notation for clarity \([\phi|_H](h)\).

Further, we can see that follwoing observatons.

• By definition, \(\forall h\in H,\phi|_H(h) = \phi(h)\)
• The restriction \(\phi|_H\) is a homomorphism (Since \(\phi\) is homomorphism).
• The kernel of \(\phi|_H\) is the intersection of the kernel of \(\phi\) with \(H\): \[ker(\phi|_H) = (ker\phi) \cap H\] There is no need to prove this. This is trivial by definition of kernel.
• Image of \(\phi|_H\) is the same as the image \(\phi(H)\) of \(H\) under the map \(\phi\). \[Im(\phi_H)=\phi(H)\]
• If \(|H|\) and \(|G'|\) have no common factor, \(\phi(H) = \{1\}\), so \(H\) is contained in the kernel. (Since, by Artin’s book corollary 2.8.13,

\[\begin{eqnarray} |Im(\phi_H)|&\biggr |&|H|\\ |Im(\phi_H)|&\biggr |& |G'| \end{eqnarray}\], Thus, if \(|H|\) and \(|G'|\) have no common factors,\(|Im(\phi_H)|=|\phi(H)| =1.\) So, \(Im(\phi_H)=\phi(H)= \{1\}\)

Now let’s see an example.

Define sign homorphism \(\sigma: S_n \to \{\pm 1\}\) by \(\sigma(x)=1\) if \(x\) is even, and \(\sigma (x)=-1\) if \(x\) is odd. Then the image of the sign homomorphism is, \[Im(\sigma)=\{\pm 1\}\] it has order \(2\).

Let \(H=\{x \in S_n : \text{if $x$ has odd order}\}\). Then \(H\) is a subgroup. So,, \(H \subset \ker(\sigma)\) (We can easily verify this.)

Furthur, The sub group of \(S_n\) with even permutations is called Alternating group (\(A_n\)).

PROBLEM

Let \(\phi:G \to G^{\prime}\) be a homomorphism with kernel \(K\), and let \(H^\prime\leq G^\prime\). Denote the inverse image \(\phi^{-1}(H^\prime)\) by \(H\). i.e.:

\[H=\phi^{-1}(H^\prime)=\{x\in G: \phi(x)\in H^\prime\}\] Then,

• \(\phi^{-1}(H^\prime)\) is a subgroup of \(G\) that contains \(K\).
• If \(H\) is a normal subgroup of \(G^\prime\), then \(\phi^{-1}(H^\prime)\) is also a normal subgroup of \(G\).
• If \(\phi\) is surjective and \(H\) is a normal subgroup of \(G\), then \(\phi^{-1}(H^\prime)\) is a normal subgroup of \(G^\prime\).

\(\phi^{-1}\) is not a map.

• Claim 1: \(K\subseteq \phi^{-1}(H^\prime)=H\)
  Let \(x\in K\). Then \(\phi(x)=1_{G^\prime}\). Since \(1_{G^\prime}\in H^\prime\). Thus, \(x\in \phi^{-1}(H^\prime)=H\). Therfore, \(K\subseteq \phi^{-1}(H^\prime)=H\)
• Claim 2: \(\phi^{-1}(H^\prime)\) is a subgroup of \(G\)
  
  • Closure :Suppose \(x,y\in H= \phi^{-1}(H^\prime)\). Then \(\phi(x),\phi(y)\in \phi(H^\prime)\). Since \(\phi\) is hormophism \(\phi(x)\phi(y)=\phi(xy)\). Since \(H^\prime\leq G^\prime\), then \(\phi(xy)=\phi(x)\phi(y)\in H^\prime\). Thus, \(xy\in \phi^{-1}(H^\prime)=H\)
  • Identity : Since \(\phi(1_G)=1_{G^\prime}\), \(1_G\in \phi^{-1}(H^\prime)=H\)
  • Inverse : Let \(x\in \phi^{-1}(H^\prime)=H\). Then \(\phi(x)\in H^\prime\) and since \(H^\prime \leq G^\prime\),then \((\phi(x))^{-1}\in H^\prime\). Since \(\phi\) is a homomorphism, \((\phi(x))^{-1}=\phi(x^{-1})\), Thus \((\phi(x))^{-1}=\phi(x^{-1})\in H^\prime\). Hence, \(x^{-1}\in \phi^{-1}(H^\prime)\).
• Claim 3: If \(H^\prime\) is a normal subgroup of \(G^\prime\), then \(\phi^{-1}(H^\prime)\) is also a normal subgroup of \(G\).
  Now suppose that \(H^\prime\) is a normal subgroup of \(G^\prime\). Let \(x\in H\) and \(g\in G\). Then, since \(\phi\) is hormorphism, \[\phi(-1)=\phi(g)\phi(x)\phi(g^{-1})=\phi(g)\phi(x)(\phi(g))^{-1}\] So, \(\phi(gxg^{-1})\) is conjugate of \(\phi(x)\). Since \(x\in H=\phi^{-1}(H^\prime)\),the \(\phi(x)\in H^\prime\). Thus, \(\phi(gxg^{-1})=\phi(g)\phi(x)(\phi(g))^{-1}\in H^\prime\). Hence, \(gxg^{-1}\in \phi^{-1}(H^\prime)=H\).
• Claim 4: If \(\phi\) is surjective and \(H\) is a normal subgroup of \(G\), then \(\phi^{-1}(H^\prime)\) is a normal subgroup of \(G^\prime\).
  Suppose that \(\phi\) is surjective and \(H\) is a normal subgroup of \(G\). Let \(a\in H^\prime\) and \(b\in G^\prime\). Since \(\phi\) is surjcetive, There are elements \(x\in H\) and \(y\in G\) such that \(\phi(x) = a\) and \(\phi(y) = b\). Since \(H\) is normal \(yxy^{- 1}\in H\), thus \(\phi(yxy^{-1})=bab^{-1}\in H^\prime\).

let denote the determinant homomorphism \[det:GL_n(\mathbb{R})\to \mathbb{R}^\times\]. Note that \(\mathbb{R}^+\unlhd \mathbb{R}^\times\). (It is very clear that $The set of positive real numbers is a subgroup of \(\mathbb{R}^\times\), and since \(\mathbb{R}^\times\) is abliean, \(\mathbb{R}^+\) is normal.) Now consider the inverse image of \(\mathbb{R}^+\), \[det^{-1}(\mathbb{R}^+)=\{A\in GL_n(\mathbb{R}):det(A)\in \mathbb{R}^+\}=SL_n(\mathbb{R})\]. By above proposition we can ge that \(SL_n(\mathbb{R})\) is normal subgroup. (Because \(\mathbb{R}^+\unlhd \mathbb{R}^\times\) and \(det\) is surjective map.)

Let \(\phi: G \to G^\prime\) be a surjective group homomorphism with kernel \(K\). There is a bijective correspondence between subgroups of \(G^\prime\) and subgroups of \(G\) that contain \(K\): \[\{\text{subgroups of $G$ that contain $K$}\}\longleftrightarrow \{\text{subgroups of $G^\prime$}\}\] This corresponde nce is defined as follows: \[\begin{aligned} \text{a subgroup $H$ of $G$ that contains K} &\rightsquigarrow \text{ its image} (\phi(H) \in G^\prime,\\ \text{a subgroup $H^\prime$ of $G^\prime$} &\rightsquigarrow \text{ its inverse image $\phi^{-1}(H^\prime)$ in $G$}. \end{aligned}\]

• If \(H\) and \(H^\prime\) are corresponding subgroups, then \(H\) is normal in \(G\) if and only if is normal in \(G^\prime\).
• If If \(H\) and \(H^\prime\) are corresponding subgroups, then \[|H|= |H^\prime||K |\]

Let \(H\) be subgroup of \(G\) that contain \(K\). Let \(H^\prime\) be a subgroup of \(G^\prime\). Now we need to check folllwings,

• \(\phi(H)\) is a subgroup of \(G^\prime\).

• \(\phi^{-1}(H^\prime)\) is a subgroup of \(G\), and it contains \(K\).

• \(H^\prime\) is a normal subgroup of \(G^\prime\) if and only if \(\phi^{-1}(H^\prime)\) is a normal subgroup of \(G\).

• (bijectivity of the correspondence) \(\phi(\phi^{-1}(H^\prime)) =H^\prime\) and \(\phi^{-1}(\phi(H))=H\)

• \(|(\phi^{-1} (H^\prime)| = |H^\prime||K|\).

• Claim 1: \(\phi(H)\) is a subgroup of \(G^\prime\).
  

  • Closure: Let $\(x,y\in phi(H)\). Then there is \(a,b\in H\) such that \(\phi(a)=x\) and \(\phi(b)=y\). Since \(\phi\) is hormorphism, \(xy=\phi(a)\phi(b)=\phi(ab)\). Sine \(H\leq G\), \(ab\in H,~ xy=\phi(ab)\in \phi(H)\).
  • Identity : Since \(1_G\in H\), \(\phi(1_G)=1_{G^\prime}\in \phi(H)\)
  • Inverse : Let \(x\in \phi(H)\). Then there exist \(a\in H\) such that \(\phi(a)=x\). Since \(H\leq G\), \(a^{-1}\) exists in \(H\). \(\phi(a^{-1})=\phi(a)^{-1}=x^{-1}\in \phi(G)\).

• Claim 2: \(\phi^{-1}(H^\prime)\) is a subgroup of \(G\), and it contains \(K\).
  This is true from proportion @ref(prp:2104)

• Claim 3: \(H^\prime\) is a normal subgroup of \(G^\prime\) if and only if \(\phi^{-1}(H^\prime)\) is a normal subgroup of \(G\).
  Alreday this prooved in proportion @ref(prp:2104)

• Claim 4.1: \(\phi(\phi^{-1}(H^\prime))=H^\prime\)

  • \(\phi(\phi^{-1}(H^\prime))\subset H^\prime\)
    Lett \(x\in \phi(\phi^{-1}(H^\prime))\). Then there exist \(y\in\phi^{-1}(H^\prime)\) such that \(\phi(y)=x\). Then by definiton of the pre image \(x=\phi(y)\in H^\prime\).

  • \(\phi(\phi^{-1}(H^\prime))\supset H^\prime\)
    Let \(a\in H^\prime\). Since \(\phi\) is surjective, there exists \(b\in G\) such that \(\phi(b)=a\in H^\prime\). Thus \(b\in \phi^{-1}(H^\prime)\). Hence \(a=\phi(b)=\phi(\phi^{-1}(H^\prime))\).

Therefore, \(\phi(\phi^{-1}(H^\prime))=H^\prime\).

• Claim 4.2: \(\phi^{-1}(\phi(H))=H\)
  • \(\phi^{-1}(\phi(H))\supset H\)
    Let \(x\in H\). Then \(\phi(x) \in \phi(H)\). So, \(x\in \phi^{-1}(\phi(H))\).
  • \(\phi^{-1}(\phi(H))subset H\)
    Let \(y\in \phi^{-1}(\phi(H))\). By definition of inverse image, \(\phi(y)\in \phi(H)\). Then there exist \(z\in H\) such that \(\phi(y)=\phi(z)\). Then \(z^{-1}y\) is in the kernal \(K\).(by Artins book proposition 2.5.8). Since \(K \subset H\), \(z^{-1}y\in H\). So, \(a\in H\) and \(a^{-1}z\in H\). Thus, \(a(a^{-1}x)=x\in H\). Hence \(\phi^{-1}(\phi(H))\subset H\)

Therefore, \(\phi^{-1}(\phi(H))=H\) - Claim 5 : \(|(\phi^{-1} (H^\prime)| = |H^\prime||K|\).

Problem

Recall

\[\begin{aligned} S_4 &= \left\{ id, (1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 3 4), (1 2 3 4) \right\}\\ S_3 &= \left\{ id, (1 2), (1 3), (2 3), (1 2 3), (1 3 2) \right\} \end{aligned}\]

There are 6 such subgroups of \(S_3\),\[ <(1 2)>, <(1 3)>, <(2 3)>, <(1 2 3)>= <(1 3 2)>,S_3\]. There is one proper subgroup of order 3. That is \(<(123)>\). There are 3 subgroups of order 2. They are \(<(1 2)>, <(1 3)>, <(2 3)>\)

The Correspondence Theorem tells us that there are four proper subgroups of \(S_4\) that contain \(K\).

Problem

Product Groups

Let \(G,G'\) be two groups. The product set \(G \times G'\), the set of pairs of elements \((a, a')\) with a in G and a’ in G’, can be made into a group by component-wise multiplication,

multiplication of pairs is defined by the rule,

\[(a,a')\cdot(b,b')=(ab,a'b')\quad \text{for } a,b\in G \text{ and for } a',b'\in G\] Let’s prove that \(G\times G'\) is a group.

Let \(G,G'\) be two groups and let \(a,b,c\in G\) and \(a',b',c'\in G'\)

• Closure : \((a,a')\cdot (b,b')=(ab,a'b')\). So, since \(a,b\in G\) and \(a',b'\in G'\) and \(G\) and \(G'\) be a group, then \(ab\in G\) and \(a'b'\in G\). Thus, \((a,a')\cdot(b,b')=(ab,a'b')\in G \times G'\).
• Asscitivity: We can obtain following using asscivity property of group \(G\) and \(G'\). \[((a,a')\cdot(b,b'))\cdot(c,c')=(ab,a'b')\cdot(c,c') =(abc,a'b'c')= (a,a')\cdot(bc,b'c')=(a,a')\cdot((b,b')\cdot(c,c'))\]
• identity: (1_G,1_{G’}) is the idenitity
  \[(1_G,1_{G'})\cdot(b,b')=(1_Gb,1_{G'}b')=(b,b') =(b1_G,b'1_{G'})=(b,b')\cdot(1_G,1_{G'})\]
• Inverse : The inverse of \((a,a')\) is \((a^{-1},(a')^{-1})\)
  \[(a,a')\cdot(a^{-1},(a')^{-1})=(aa^{-1},a'(a')^{-1})=(1_G,1_{G'})=(a^{-1}a,(a')^{-1}a')=(a^{-1},(a')^{-1})\cdot(a,a')\]

So, The group obtained in this way is called the product of \(G\) and \(G'\).

It is related to the two factors \(G\) and \(G'\) in a simple way that we can sum up in terms of some homomorphisms. The homomorphisms are defined as follows, \[\begin{eqnarray} i:G &\rightarrow& G \times G' \\ x &\mapsto& (x,1)\\\\ i':G' &\rightarrow& G \times G' \\ x &\mapsto& (x',1)\\\\ p:G \times G' &\rightarrow& G' \\ (x,x') &\mapsto& x\\\\ p':G' \times G' &\rightarrow& G' \\ (x,x') &\mapsto& x'\\\\ \end{eqnarray}\]

Observe that \(i\) and \(i'\) are injective and \[Im(i)=G\times 1_G'\leq G\times G' \text{ and }Im(i')=1_G\times G'\leq G\times G'\] The maps \(p\) and \(p '\) are called projections and they are surjective. \[\ker(p)=1\times G'\text{ and }\ker(p')=G\times 1_{G'}\] It is obviously desirable to decompose a given group \(G\) as a product, that is, to find investigate groups \(H\) and \(H '\) such that \(G\) is isomorphic to the product \(H \times H'\). The groups \(H\) and \(H'\) will be simpler, and the relation between \(H\times H'\) and its factors is easily understood. It is rare that a group is a product, but it does happen occasionally

Consider a cyclicgroup of order \(6\) can be decomposed. It migheted be suprised you. A cyclic group \(C_6\) of order \(6\) isisomorphic to the product \(C_2 \times C_3\) of cyclic groups of orders \(2\) and \(3\). \[C_6\equiv C_2 \times C_3\] Let say

\[\begin{eqnarray} C_2&=&<y>=\{1,y\},\text{ with } y^2=1\\ C_3&=&<z>=\{1,z,z^2\}, \text{ with } y^3=1 \end{eqnarray}\]

Let \(x\in C_2\times c_3\). Then there exist \(p\in C_2\) and \(q\in C_3\) such that \(x=(p, q)\). Let’s find order of \(x\), that is the smallest positive integer \(k\) such that \(x^k =(y^k,z^k)\) is the identity \((1, 1)\).

• Case-I: \(k=1, x^1=(y^1,z^1)=(y,z)\)
• Case-II: \(k=2, x^2=(y^2,z^2)=(1,z^2)\)
• Case-III: \(k=3, x^3=(y^3,z^3)=(y,1)\)
• Case-IV: \(k=4, x^4=(y^4,z^4)=(1,z)\)
• Case-V: \(k=5, x^5=(y^5,z^5)=(1,z^2)\)
• Case-VI: \(k=6, x^5=(y^6,z^6)=(1,1)\)

Thus, the smallest positive integer \(k\) such that \(x^k=(1, 1)\) is 6. So, order of \(x\) is \(6\). Since \(C_2\times C_3\) has order \(6\). Furthur, \[C_2\times C_3 =<x>\] The powers of \(x\) are \[(1,1),(y,z),(1,z^2),(y,1),(1,z),(y,z^2)\]

Here \(i=0,1,2...5\)

So let’s try to see above result more generally.

Let \(r\) and \(s\) be relatively prime integers. A cyclic group of order \(rs\) is isomorphic to the product of a cyclic group of order \(r\) and a cyclic group of order \(s\).

On the other hand, a cyclic group of order \(4\) is not isomorphic to a product of two cyclic groups of order \(2\). \[C_4\not\equiv C2 \times C_2\] Because , every element of \(C_2 \times C_2\) has order \(1\) or \(2\),whereasa cyclic group of order \(4\) contains two elements of order \(4\).

The next proposition describes product groups.

Let \(G\) be a group and let \(H,K\leq G\), and let \(f: H \times K \rightarrow G\) be the multiplication map, defined by \(f(h, k) = hk\). Its image is the set \(HK = \{hk | h \in H, k \in K\}\).

1. \(f\) is injective if and only if \(H \cap K = \{1\}\).
2. \(f\) is a homomorphism from the product group \(H \times K\) to \(G\) if and only if elements of \(K\) commute with elements of \(H\): \(hk = kh\).
3. If \(H\) is a normal subgroup of \(G\), then \(HK\) is a subgroup of \(G\).
4. \(f\) is an isomorphism from the product group \(H \times K\) to \(G\) if and only if \(H \cap K = \{1\}\), \(HK = G\), and also \(H\) and \(K\) are normal subgroups of \(G\).

It is important to note that the multiplication map may be bijective though it isn’t a group homomorphism. This happens, for instance, when \(G = S_3\), and with the usual notation, \(H = \langle x \rangle\) and \(K = \langle y \rangle\).

• (\(\Longrightarrow\)) We are going to use proof by contrapositive. So, suppose that \(x\in H\cap K\) such that \(x \neq 1\). Since \(x\in H\) then \(x^{-1} \in H\). \[f(x^{-1},x)=x^{-1}\cdot x=1=1\cdot 1=f(1,1)\] Thus, \(f\) is not injective. Thus, if \(f\) is injective then \(H \cap K = \{1\}\). -(\(\Longleftarrow\)) Now Suppose that \(H \cap K = \{1\}\). Let \((h_1, k_1),(h_2, k_2)\in H \times K\) such that \(h_1k_1 = h_2k_2\). Now multiply both sides of this equation on the left by \(h_1^{-1}\) and on the right by \(k_2^{-1}\), \[\begin{eqnarray} h_1k_1 &=& h_2k_2\\ h_1^{-1}(h_1k_1)k_2^{-1} &=& h_1^{-1}(h_2k_2)k_2^{-1}\\ (h_1^{-1}h_1)k_1k_2^{-1} &=& h_1^{-1}h_2(k_2k_2^{-1})\\ k_1k_2^{-1} &=& h_1^{-1}h_2 \end{eqnarray}\] Note that \((k_1k_2^{-1})\in K\) and \(h_1^{-1}h_2\in H\). Since \(H \cap K = \{1\}\), \[k_1k_2^{-1} = h_1^{-1}h_2 = 1\]. Then, \[ k_1=k_2~~\text{and}~~~~h_1=h_2. \] Then \[(h_1, k_1) = (h_2, k_2)\]

2. Let \((h_1, k_1),(h_2, k_2)\in H \times K\). Now consider, \[\begin{eqnarray} f((h_1,k_1)\cdot (h_2,k_2))&=&f((h_1h_2,k_1k_2))=h_1h_2k_1k_2\\ f((h_1,k_1))\cdot f((h_2,k_2))&=& (h_1k_1)\cdot(h_2k_2)=h_1k_1h_2k_2 \end{eqnarray}\]

\[\begin{eqnarray} f\text{ is homorphism} &\iff & f((h_1,k_1)\cdot (h_2,k_2))=f((h_1,k_1))\cdot f((h_2,k_2))\\ &\iff & h_1h_2k_1k_2=h_1k_1h_2k_2\\ &\iff & h_1^{-1}(h_1h_2k_1k_2)k_2^{-1}=h_1^{-1}(h_1k_1h_2k_2)k_2^{-1}\\ &\iff & (h_1^{-1}h_1)h_2k_1(k_2k_2^{-1})=(h_1^{-1}h_1)k_1h_2(k_2k_2^{-1})\\ &\iff & (h_1^{-1}h_1)h_2k_1(k_2k_2^{-1})=(h_1^{-1}h_1)k_1h_2(k_2k_2^{-1})\\ &\iff & h_2k_1=k_1h_2 \end{eqnarray}\]

• Suppose that \(H\) is a normal sub group of G. Note that \[KH=\bigcup_{k\in K}kH~~\text{ and } ~~HK=\bigcup_{k\in K}Hk\] Since, \(H\) is normal, \(kH=Hk\) for all \(k\in K\). So, \(HK=KH\). We are going to use sub group test,

  • non-emptyness: Clearly, \(1=1\cdot 1\in HK\) (because \(1\in H\) and \(1\in K\)
  • closure: \[HKHK=HHKK=HK\] Thus, \(HK\) is closed under multiplication.
  • closed under inverse: Let \(hk\in HK\). Then \[(hk)^{-1} = k^{-1}\cdot h^{-1} \in KH=HK \] This proves closure of \(HK\) under inverses.

• \((\Longleftarrow)\): Suppose that \(H \cap K = \{1\}\), \(HK = G\), and also \(H,K \trianglelefteq G\). According to the (b), \(f\) is a homomorphism from \(H \times K\) to \(G\) if and only if \(hk = kh\) for all \(h\in H\). Consider, \[(hkh^{-1})k^{-1}=h(kh^{-1}k^{-1}).\] Note that, Since \(K \trianglelefteq G\), \(hkh^{-1}\in K\). So,\((hkh^{-1})k^{-1}\in K\). Similary we can show that \(h(kh^{-1}k^{-1})\in H\).Thus,\[(hkh^{-1})k^{-1}=h(kh^{-1}k^{-1})\in H\cap K.\] But from our hypothesis, \(H\cap K=\{1\}\). Hence,

\[\begin{eqnarray} hkh^{-1}k^{-1}&=&1\\ hk&=& kh \end{eqnarray}\]

Therefore, by (b) we can \(f\) is hormorphism. Now we have to proove bijectivity, - Injectivity: Already prooved in @ref(prp:prp2114) (a) - Surjectivity: For any \(g \in G\), we can write \(g = hk\) for some \(h \in H\) and \(k \in K\) (since \(Im(f)=HK = G\)). Thus, \(f(h, k) = hk = g\), so \(f\) is surjective.

Therefore \(f\) is an isomorphism.

• \((\Longrightarrow)\) Now suppose that \(f\) is isomorphism.

  • Claim i: \(H \cap K = \{1\}\).
    Suppose that \(x\in H \cap K\). Then, \[f(1,x)=x=f(x,1)\] Since \(f\) is an isomorphism (and hence injective), we must have \(x = 1\). Therefore, \(H \cap K = \{1\}\).

  • Claim ii :\(HK = G\).
    Since \(f\) is surjective, for any \(g \in G\), there exist \(h \in H\) and \(k \in K\) such that \(f(h, k) = hk = g\). Therefore, \(HK = G\).

  • Claim iii : \(H\) and \(K\) are normal in \(G\).
    For any \(h \in H\) and \(g \in G\), we can write \(g = hk\) for some \(k \in K\). Then \[ghg^{-1} = (hk)h(hk)^{-1} =hk(hh^{-1})k^{-1}=h(kk^{-1})=h\in H.\] Thus, \(H\trianglelefteq G\). Similarly, we can show that \(K\trianglelefteq G\).

There are two isomorphism classes of groups of order \(4\), the class of the cyclic group \(C_4\) of order \(4\) and the class of the Klein Four Group,which is isomorphic to the product \(C_2 \times C_2\) of two groups of order \(2\).

Let \(G\) be a group of order 4. The order of any element \(x\) of \(G\) divides 4, so there are two cases to consider:

• Case 1: \(G\) contains an element of order \(4 (|G|=4)\). Then \(G\) is a cyclic group of order \(4\).
• Case 2: Every element of \(G\) except the identity has order \(2\).
  In this case, \(x = x^{-1}\) for every element \(x\) of \(G\). Let \(x\) and \(y\) be two elements of \(G\). Then \(xy\) has order \(2\), so \((xy)(x^{-1}y^{-1}) = (xy)(xy) = 1\). This shows that \(x\) and \(y\) commute,

\[\begin{eqnarray} xyx^{-1}y^{-1} &=& 1\\ xyx^{-1}(y^{-1} y) &=& y\\ xyx^{-1} &=& y\\ xy(x^{-1}x) &=& yx\\ xy &=& yx \end{eqnarray}\]

since \(x,y\) are arbitrary elements, \(G\) is abelian. So every subgroup is normal. We choose distinct elements \(x\) and \(y\) in \(G\), and we let \(H\) and \(K\) be the cyclic groups of order 2 that they generate. Proposition @ref(prp:prp2114) (d) shows that \(G\) is isomorphic to the product group \(H \times K\).

Quotient Groups

The set of cosets of a normal subgroup \(N\) of a group \(G\) is often denoted by \(G / N\). \[G / N \text{ is the set of cosets of $N$ in} G\]

Notation: When we regard a coset \(C\) as an element of the set of cosets, the bracket notation \([C]\) may be used. If \(C = aN\), we may also use the bar notation to denote the element \([C]\) by \(\overline{a}\), and then we would denote the set of cosets by \(\overline{G}\): \[\overline{G} = G /N\]

Let \(N\) be a normal subgroup of a group \(G\), and let \(G\) denote the set of cosets of \(N\) in \(G\). There is a law of composition on \(G\) that makes this set into a group, such that the map \(\pi: G \to \overline{G}\) defined by \(\pi(a) = \overline{a}\) is a surjective homomorphism whose kernel is \(N\).

The map is often referred to as the canonical map from \(G\) to \(\overline{G}\). The word “canonical” indicates that this is the only map that we might reasonably be talking about

The next corollary is very simple, but it is important enough to single out:

Let \(N \trianglelefteq G\), and let \(\overline{G}\) denote the set of cosets of \(N\) in \(G\). Let \(\pi: G \to \overline{G}=G/N\) be the canonical homomorphism. Let \(a_1, \ldots, a_k\in G\) such that the product \(a_1 \cdots a_k\in N\). Then \(\overline{a_1} \cdots \overline{a_k} = \overline{1}\).

Let \(p=a_1 \cdots a_k\in N\). This implies \(\pi(p)=\overline{p}=\overline{1}\).
Since is \(pi\) a homomorphism, \(\pi(p)=\pi(a_1 \cdots a_k)=\pi(a_1)\cdots \pi(a_k)=\overline{a_1}\cdots\overline{a_n}\)

Proof of @ref(thm:2122) There are several things to be done. We must

• define a law of composition on \(\overline{G}\),
• prove that the law makes \(\overline{G}\) into a group,
• prove that \(\pi\) is a surjective homomorphism, and
• prove that the kernel of \(\pi\) is \(N\).

If \(A,B \subseteq G\) then \(AB\) denotes the set of products \(ab\): \[AB:=\{x\in G: x=ab, a\in A, b\in B\}\]

• We will call this a product set, though in some other contexts the phrase “product set” refers to the set \(A \times B\) of pairs of elements

Let \(N\) be a normal subgroup of a group \(G\), and let \(a N\) and \(b N\) be cosets of \(N\). The product set \((aN )(bN )\) is also a coset and \[(aN )(bN )=\{x\in G: x=anbn' ~\& ~ n,n'\in N\}=abN\].

Since \(N\) is a subgroup, \(N N = N\).
Since \(N\) is normal, left and right cosets are equal: \(N b = bN\). \[(aN )(bN ) = a (N b )N = a (b N )N = a b N N = abN\]

• This lemma allows us to define multiplication on the set \(\overline{G} = G / N\).
• Using the bracket notation, the definition is this: If \(C_1\) and \(C_2\) are cosets, then \([C_1][C_2] = [C_1 C_2]\), Where \(C_1 C_2\) is the product set.
• The lemma shows that this product set is another coset.
  
• To compute the product \([C_1][C_2]\), take any elements \(a \in C_1\) and \(b \in C_2\). Then \(C_1 = aN\) , \(C_2 = bN\) , and \(C_1 C_2\) is the coset \(a b N\) that contains \(ab\). So, \[[aN][bN] = [abN] \text{ or } \overline{a}\overline{b} = \overline{ab}.\] Then by definition of the map \(\pi\) in theorm @ref(thm:2122), \[\pi(ab)=\overline{a}\overline{b} = \overline{ab}=\pi(a)\pi(b)\] The fact that \(\pi\) is a homomorphism will follow, once we show that G is a group. Since the canonical map \(\pi\) is surjective , the next lemma proves this

Let \(G\) be a group, and let \(Y\) be a set with a law of composition (both laws written with multiplicative notation). Consider a surjective map \(\varphi : G \to Y\) with the homomorphism property: \(\varphi(ab) = \varphi(a)\varphi(b)\) for all \(a\) and \(b\) in \(G\). Then \(Y\) is a group, and \(\varphi\) is a homomorphism.

The group axioms that are true in \(G\) are carried over to \(Y\) by the surjective map \(\varphi\).

• Closure : Let \(y_1,y_2\in Y\). Since \(\varphi\) is surjective, \(y_1=\varphi(x_1),y_2=\varphi(x_2)\) for some \(x_1,x_2\in G\). \[y_1y_2=\varphi(x_1)\varphi(x_2)=\varphi(x_1x_2)\in Y\]

• Assciativity Property :
  Let \(y_1\), \(y_2\), and \(y_3\) be elements of \(Y\). Since \(\varphi\) is surjective, \(y_i = \varphi(x_i)\) for some \(x_i\) in \(G\). Then

\[\begin{eqnarray} (y_1 y_2) y_3 &=& (\varphi(x_1) \varphi(x_2)) \varphi(x_3) = \varphi(x_1 x_2) \varphi(x_3) = \varphi((x_1 x_2) x_3) \\ &=^*& \varphi(x_1 (x_2 x_3)) = \varphi(x_1) \varphi(x_2 x_3) = y_1 (y_2 y_3) \end{eqnarray}\]

The equality marked with an asterisk \((=^*)\) is the associative law in \(G\). The other equalities follow from the homomorphism property of \(\varphi\).

• Identity Let \(y\in Y\). Since \(\varphi\) is surjective, there exist \(x\in G\) such that \(y=\varphi(x)\). Let \(1_G\) be identity of \(G\) Then, \[y\varphi(1_G)=\varphi(x)\varphi(1_G)=\varphi(x1_G)=\varphi(x)=y= \varphi(1_Gx)=\varphi(1_G)\varphi(x)=\varphi(1_G)y\] Thus, \(\varphi(1_G)\) is the identity of \(Y\).

• Inverse : Let \(y\in G\). Since \(\varphi\) is surjective, there exist \(x\in G\) such that \(y=\varphi(x)\), \[y\varphi(x^{-1}=\varphi(x)\varphi(x^{-1}) =\varphi(xx^{-1})=\varphi(1_G)=\varphi(x^{-1}x)=\varphi(x^{-1})\varphi(x)=\varphi(x^{-1})y\] Thus, \(y^{-1}=\varphi(x)\)

The only thing remaining to be verified is that the kernel of the homomorphism \(\pi\) is the subgroup \(N\).

\[\begin{eqnarray} \pi (a) =\pi(1) &\iff & \bar{a}= \bar{1}\\ & \iff & [aN] = [1N]\\ & \iff & a \in N. \end{eqnarray}\] Thus, \(\ker(\pi)=N\)

A Schematic Diagram of Coset Multiplication.

Our assumption that \(N\) is a normal subgroup of \(G\) is crucial to lemma @ref(lem:2125). If \(H\) is not normal, there will be left cosets \(C_1\) and \(C_2\) of \(H\) in \(G\) such that the product set \(C_1 C_2\) does not lie in a single left cosets.

Let’s see an example for this. Going back once more to the subgroup \(H = \langle y \rangle\) of \(S_3\). Note that the subgroup \(H\) is not normal.
The product set \((1H)(xH)\) contains four elements: \[(1H)(xH)=\{1, y\}\{x, xy\} = \{x, xy, x^2y, x^2\}\]. It is not a coset.

The next theorem relates the quotient group construction to a general group homomorphism, and it provides a fundamental method of identifying quotient groups.

Let \(\varphi: G \to G'\) be a surjective group homomorphism with kernel \(N\). The quotient group \(\overline{G}=G/N\) is isomorphic to the image \(G'\). To be precise, let \(\pi: G \to \overline{G}=G/N\) be the canonical map. There exists a unique isomorphism \(\overline{\varphi}: G/N \to G'\) such that \(\varphi = \psi \circ \pi\).

The elements of \(\overline{G}=G/N\) are the cosets of \(N\), and they are also the fibres of the map \(\varphi\). The map \(\overline{\varphi}\) referred to in the theorem is the one that sends a non-empty fibre to its image: \(\overline{\varphi}(\overline{x})=\varphi(x)\). For any surjective map of sets \(\varphi: G \to G'\), one can form the set \(\overline{G}\) of fibres, and then one obtains a diagram as above, in which \(\overline{\phi}\) is the bijective map that sends a fibre to its image. When \(\phi\) is a group homomorphism, \(\overline{\phi}\) is an isomorphism because \[\overline{\phi}(\overline{ab}) = \phi(ab) = \phi(a)\phi(b) = \overline{\phi}(\overline{a})\overline{\phi}(\overline{b}).\]

Let \(\varphi: G \rightarrow G'\) be a group homomorphism with kernel \(N\) and image \(H'\). The quotient group \(\overline{G}=G/N\) is isomorphic to the image \(H'\).

Vector Space

Subspace of \(\mathbb{R}^n\)

Vector spaces, which are the focus of this chapter, can be thought of as subsets within the realm of \(\mathbb{R}^n\), where \(n\) represents the number of dimensions in real-numbered vector space. This topic is explored further in the current section, with a formal definition provided in Section 3.3.

While row vectors are more compact, column vectors are preferred due to their compatibility with matrix multiplication. For efficiency in written form, we often represent a column vector as \((a_1, \ldots, a_n)^t\) using the transpose operation. As established in Chapter 1, we treat a column vector and a point in \(\mathbb{R}^n\) with identical coordinates as equivalent. Typically, column vectors are indicated by lower-case letters like \(v\) or \(w\). When \(v\) corresponds to \((a_1, \ldots, a_n)^t\), this particular representation is referred to as the coordinate vector of \(v\).

We consider two operations on vectors: \[\begin{align*} & \text{vector addition: } & \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} &+ \begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix} &= \begin{bmatrix} a_1 + b_1 \\ \vdots \\ a_n + b_n \end{bmatrix}, \\ & \text{scalar multiplication: } & c &\cdot \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} &= \begin{bmatrix} ca_1 \\ \vdots \\ ca_n \end{bmatrix}. \end{align*}\] These operations make \(\mathbb{R}^n\) into a vector space.

A subset \(W\) of \(\mathbb{R}^n\) is a subspace if it has these properties:

1. If \(w\) and \(w'\) are in \(W\), then \(w + w'\) is in \(W\).
2. If \(w\) is in \(W\) and \(c\) is in \(\mathbb{R}\), then \(cw\) is in \(W\).
3. The zero vector is in \(W\).

There is another way to state the conditions for a subspace,

Let \(W\) be a non-empty set. If \(w_1, w_2, \ldots, w_n\) are elements of \(W\) and \(c_1, c_2, \ldots, c_n\) are scalars, then the linear combination \(c_1w_1 + c_2w_2 + \ldots + c_nw_n\) is also in \(W\).

Given an \(m \times n\) matrix \(A\) with coefficients in \(\mathbb{R}\), the set of vectors in \(\mathbb{R}^n\) whose coordinate vectors solve the homogeneous equation \(AX = 0\) is called the nullspace of \(A\).

(Example for subspaces) Systems of homogeneous linear equations.

Given an \(m \times n\) matrix \(A\) with coefficients in \(\mathbb{R}\), the set of vectors in \(\mathbb{R}^n\) whose coordinate vectors solve the homogeneous equation \(AX = 0\) is a subspace, called the nullspace of \(A\). Though this is very simple, we will check the conditions for a subspace:

• Suppose that \(X\) and \(Y\) be solutions of given homogenous sysmstem. Then \(AX=0\) and \(AY=0\). Thus, \[A(X+Y)=AX+AY=0+0=0\] Thus, \((X+Y)\) is a solution.
• If \(X\) and \(Y\) be solutions of given homogenous sysmstem. Then \(AX=0\). Let \(c\in \mathbb{R}\). \[A(cX)=c(AX)=c\cdot 0=0\] Thus, \(cX\) is a solution.
• \(A0=0\). Thus, the zero vector \(0\) is solution.

Therfore nullspace of \(A\) is a subspace.

The zerospace \(W = \{0\}\) and the whole space \(W = \mathbb{R}^n\) are subspaces.

A subspace is called a proper subspace if it’s not the entire space

The next proposition describes the proper subspaces of \(\mathbb{R}^2\).

Let \(W\) be a proper subspace of the space \(\mathbb{R}^2\), and let \(w\) be a non-zero vector in \(W\). Then \(W\) consists of the scalar multiples \(cw\) of \(w\). Distinct proper subspaces have only the zero vector in common.

The subspace consisting of the scalar multiples \(cw\) of a given nonzero vector \(w\) is called the subs pace spanned by \(w\).
Geometrically, it is a line through the origin in the plane \(\mathbb{R}^2\).

(Proof of the proportion) Redo it.

Fields

A field \(F\) is a set together with two laws of composition:

1. Addition: \[\begin{eqnarray} +:F \times F & \rightarrow & F \\ (a,b) & \mapsto & a+b \end{eqnarray}\]
2. Multiplication \[\begin{eqnarray} \times:F \times F & \rightarrow & F \\ (a,b) & \mapsto & ab \end{eqnarray}\]

which satisfy these axioms:

• (F1) Addition makes \(F\) into an abelian group \((F,+)\);
  its identity element is denoted by \(0\).

• (F2) Multiplication is commutative, and it makes the set of nonzero elements of \(F\) into an abelian group \(F^{\times}\); its identity element is denoted by \(1\).

• (F3) Distributive law: \[\text{For all } a, b,\text{ and } c \in F,~ a(b + c) = ab + ac\].

The first two axioms describe properties of the two laws of composition, addition and multiplication, separately. The third axiom, the distributive law, relates the two laws.

we will be familiar with the fact that the real numbers satisfy these axioms, but the fact that they are the only ones needed for the usual algebraic operations can only be understood after some experience.

The next lemma explains how the zero element multiplies.

Let \(F\) be a field.

1. The elements 0 and 1 of \(F\) are distinct.

2. For all \(a\) in \(F\), \(a0 = 0\) and \(0a = 0\).

3. Multiplication in \(F\) is associative, and 1 is an identity element.

1. Axiom (ii) implies that 1 is not equal to 0.

2. Since 0 is the identity for addition, \(0 + 0 = 0\). Then \(a0 + a0 = a(0 + 0)= a0\). Since \((F,+)\) is a group, we can cancel \(a0\) to obtain \(a0 = 0\), and then \(0a = 0\) as well.

3. Since \(F \setminus \{0\}\) is an abelian group, multiplication is associative when restricted to this subset. We need to show that \(a(bc) = (ab)c\) when at least one of the elements is zero. In that case, (b) shows that the products in question are equal to zero. Finally, the element 1 is an identity on \(F \setminus \{0\}\). Setting \(a = 1\) in (b) shows that 1 is an identity on all of \(F\).

• Real numbers (\(\mathbb{R}\))
  Easy to verify
• Complex Numbers (\(\mathbb{C}\))
  Easy to verify
• Prime field \[\mathbb{F}_p:={\overline{0},\overline{1},...,\overline{p-1}}=\mathbb{Z}/p\mathbb{Z}\] We saw in the previous chapter that the set \(\mathbb{Z}/n\mathbb{Z}\) of congruence classes modulo an integer \(n\) has laws of addition and multiplication derived from addition and multiplication of integers. All of the axioms for a field hold for the integers, except for the existence of multiplicative inverses. And as noted in previous chapters, such axioms carry over to addition and multiplication of congruence classes. But the integers are not closed under division, so there is no reason to suppose that congruence classes have multiplicative inverses. The class of 2, for example, has no multiplicative inverse modulo \(6\). It is somewhat surprising that when \(p\) is a prime integer, all nonzero congruence classes modulo \(p\) have inverses, and therefore the set \(\mathbb{Z}/ p \mathbb{Z}\) is a field. In the next theorem, we prove the existence of multiplicative inverse of non-zero element in prime field.

Let \(p\) be a prime integer. Every nonzero congruence class modulo \(p\) has a multiplicative inverse, and therefore \(\mathbb{F}_p\) is a field of order \(p\).

We discuss the theorem before giving the proof.

Exercises

Chapter 1

Chapter 2

Laws of composition

Let \(S\) be a set. Prove that the law of composition defined by \(ab = a\) for all \(a\) and \(b\) in \(S\) is associative? For which sets does this law have an identity?

Solution: Let \(a,b,c\in S\). Now consider following \[(ab)c=(ac)=a=(ab)=a(bc)\] Thus, the given law of composition is associative.

If the given law of composition has an identity element whenever every element \(a\in S\) has a multiplicative inverse./ In other words, for every element \(a\in S\), there exists an element \(e \in S\) such that \[ae = ea = a.\] Thus, \(ae=a=ea=e\). So, the identity element is the same as every element in S, and the law has an identity for all sets S. Thus, only singletons sets have this given law of compositions have identity.

Prove the properties of inversesthat are listed near the end ofthe section.

• If an element \(a\) has both a left inverse \(l\) and a right inverse \(r\), then \(r = l\), \(a\) is invertible and \(r\) is its inverse.

Since \(l\) is a left inverse for \(a\), then \(la = 1\). In the same way, since \(r\) is a right inverse for \(a\) the equality \(ar = 1\) holds. Let us now consider the expression \(lar\). By associativity of the composition law in a group we have \(r = 1r = (la)r = lar = l(ar) = l1 = l\). This implies that \(l = r\). Since \(l = r\), it holds also that that \(ar = 1 = la = ra\) hence \(a\) is invertible and \(r\) is its inverse.

• If \(a\) is invertible, its inverse is unique.

Let \(i_1\) and \(i_2\) be inverses of \(a\). In particular \(i_1\) is a left inverse of \(a\) and \(i_2\) is a right inverse of \(a\). By point (a) \(i_1 = i_2\).

• Inverses multiply in the opposite order: if \(a\) and \(b\) are invertible, then the product \(ab\) is invertible and \((ab)^{-1} = b^{-1}a^{-1}\).

In order to show that \(ab\) is invertible, it is enough to exhibit an element that is a right and a left inverse of \(ab\). The element \(b^{-1}a^{-1}\) is a right inverse of \(ab\) since \(abb^{-1}a^{-1} = a(bb^{-1})a^{-1} = a1a^{-1} = aa^{-1} = 1\). It is a left inverse since \(b^{-1}a^{-1}ab = b^{-1}(a^{-1}a)b = b^{-11}b = b^{-1}b = 1\). This proves that \(ab\) is invertible and that \(b^{-1}a^{-1}\) is its inverse.

Groups and Subgroups

Make a multiplication table for the symmetric group \(S_3\).

Solution

\(\circ\)	e	(12)	(13)	(23)	(123)	(132)
e	e	(12)	(13)	(23)	(123)	(132)
(12)	(12)	e	(123)	(132)	(13)	(23)
(13)	(13)	(123)	e	(132)	(23)	(12)
(23)	(23)	(132)	(123)	e	(12)	(13)
(123)	(123)	(13)	(23)	(12)	(132)	e
(132)	(132)	(23)	(12)	(13)	e	(123)

Let \(S\) be a se t with an associative law of composition and with an identity element. Prove that the subset consisting of the invertible elements in \(S\) is a group.

Let \(S^*\subseteq S\) with consisting of the invertible elements in \(S\).

• identity : We are given identity element \(1 \in S\) and \((1^{-1})(1)=1 \implies 1 \in S^*\).

• Closure : So let \(a,b\in S^*\). Then \(a,b\) are invertible. Let \(a^{-1},b^{-1}\in S^*\) be inverses of \(a\) and \(b\) respectively. Now we need to check \(ab\in S^*\).
  

\[(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=b^{-1}b=e= aa^{-1}=a(bb^{-1})a^{-1}=(ab)(b^{-1}a^{-1})\]

Thus, \((ab)^{-1}=b^{-1}a^{-1}\). So, \(ab\in S^*\) Therefore, \(S^*\) is closed under the composition.

• Associativity: The associativity property inherit from the group \(S\).

Therefore, \(S^*\) is a group.

Let \(G\) be a group.Let \(x, y, z,w \in G\).

1. Solve for \(y\), given that \(xyz^{-1}w = 1\).
2. Suppose that \(xyz = 1\). Does it follow that \(yzx = 1\)? Does it follow that \(yxz = 1\)?

Solution:

1. 
   
2. 
   

- The given statement is false. \[\begin{eqnarray} xyz&=&1\\ x^{-1}(xyz)&=&x^{-1} \cdot 1\\ (x^{-1}x)(yz)&=&x^{-1}\\ 1 \cdot(yz)&=&x^{-1}\\ (yz)&=&x^{-1}\\ (yz)z^{-1}&=&x^{-1}z^{-1}\\ y(zz^{-1})&=&x^{-1}z^{-1}\\ y \cdot 1&=&x^{-1}z^{-1}\\ yx&=&x^{-1}z^{-1}x\\ yxz&=&x^{-1}z^{-1}xz\\ \end{eqnarray}\] If \(G\) is not abelian or \(z\neq x\). \(xyz=1 \nRightarrow yxz=1\).

In which of the following cases is \(H\) a subgroup of \(G\)?

1. \(G = GL_n(\mathbb{C})\) and \(H = GL_n(\mathbb{R})\).

2. \(G = \mathbb{R}^\times\) and \(H = \{1, -1\}\).

3. \(G = \mathbb{Z}^+\) and \(H\) is the set of positive integers.

4. \(G = \mathbb{R}^+\) and \(H\) is the set of positive reals.

5. \(G = GL_2(\mathbb{R})\) and \(H\) is the set of matrices \(\begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}\) with all entries equal to 0.

Solution:

• Subset ? : Since every matrix with real entries can be interpreted as a matrix with complex entries, we conclude that \(H \subset G\).

• Closure ? : For every \(A, B \in GL_n(\mathbb{R})\) we have \(AB \in GL_n(\mathbb{R})\) since Product of invertible matrices is invertible. I proved this result in chapter 1.

• Identity ? :The identity in \(G\) is \(I_n\), the identity \(n x n\) matrix. Observe that \(I_n\in H=GL_n(\mathbb{R})\). (Because every entry of \(I_n\) is 1 or 0)

• Inverse ?: \(GL_n(\mathbb{R})\) is the set of \(n \times n\) invertible matrices with real entries. So, all the elements in \(H\) is invertible.

Therefore, \(H \leq G\)

• Subset ? : This is trivial \(\{-1,1\}\subset \mathbb{R}^\times\)

• Closure ? :

\[\begin{eqnarray} 1 \times 1 &=& 1\in H\\ (-1) \times 1 &=& (-1)\in H\\ 1 \times (-1) &=& (-1)\in H\\ (-1) \times (-1) &=& -1\in H\\ \end{eqnarray}\] Thus, \(H\) is closed. - Identity ? : The identity in \(G\) is \(1\), which is also in \(H\).

• Inverse ? : \[\begin{eqnarray} 1 \times 1 = 1 &\implies & (1)^{-1}=1\in H\\ (-1) \times (-1) = 1 &\implies & (-1)^{-1}=(-1)\in H \end{eqnarray}\] x`Therefore, \(H\) is sub group of \(G\).

3. Note that \(2\in H\), but inverse of \(2\) in \(\mathbb{Z}^+\) is \((-2) \not \in H\).
   Therefore, \(H\) is not a subgroup of \(G\).

• Subset ? : This is trivial. \(H \subset \mathbb{R}^\times\)

• Closure ? : Product of positive real number is positive. Thus, \(H\) is closed.

• Identity ? : \(1\) is the identity of \(G\), which is in \(H\).

• Inverse ? : Let \(a\in H\). Then, \[a\cdot \frac{1}{a}=\frac{1}{a}\cdot a=1\] So, \(\frac{1}{a}\in H\) is the inverse of \(a\). Thus, \(H\) is closed under inverses.

5. Let \(A=\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}\) Then \(A\in H\). But observe that \(det(A)=0\implies A\) is not invertible. Thus, \(A\not\in H\). Thus, this \(H \not\subseteq G\).

Therefore, \(H\) is not a subgroup of \(G\).

In the definition of a subgroup, the identity element in \(H\) is required to be the identity of \(G\). One might require only that H have an identity element, not that it need be the same as the identity in \(G\).

• Show that if \(H\) has an identity at all, then it is the identity in \(G\).
• Show that the analogous statement is true for inverses.

Let \(G\) be a group. We define an opposite group \(G^{\circ}\) with the law of composition \(a * b\) as follows: - The underlying set is the same as \(G\), - but the law of composition is \(a * b = ba\).

Prove that \(G^{\circ}\) is a group.

Solution:

Subgroups of the Additive Group of Integers

Let \(a = 123\) and \(b = 321\). Compute \(d = gcd(a, b)\), and express \(d\) as an integer combination \(ra + bs\).

Solution:

Prove that if \(a\) and \(b\) are positive integers whose sum is a prime \(p\), their greatest common divisor is \(1\).

Solution:

1. Define the greatest common divisor of a set \(\{a_1, ..., a_n\}\) of \(n\) integers. Prove that it exists, and that it is an integer combination of \(a_1, ..., a_n\).
2. Prove that if the greatest common divisor of \(\{a_1, ..., a_n\}\) is \(d\), then the greatest common divisor of \(\{a_1/d, ..., a_n/d)\) is \(1\).

Solution:

b.

Cyclic group

Let \(a\) and \(b\) be elements of a group \(G\). Assume that a has order \(7\) and that \(a^3b = ba^3\). Prove that \(a b = ba\).

Solution:

An \(n\)th root of unity is a complex number \(z\) such that \(z^n = 1\).

1. Prove that the nth roots of unity form a cyclic subgroup of \(\mathbb{C}^\times\) of order \(n\).
2. Determine the product of all the \(n\)th roots of unity.

Solution:

Let \(a\) and \(b\) be elements of a group \(G\). Prove that \(ab\) and \(ba\) have the same order

Let \(a\) and \(b\) be elements in a group \(G\) with identity \(e\).

Suppose that \(ab\) has finite order \(n\). Then \((ab)^n = e\) and \(n\) is the smallest positive integer for which this equation is true. We also have that \[ (ba)^{n+1} = (ba)(ba) \cdots (ba) = b\underbrace{(ab) \cdots (ab)}_{n \text{ times}}a =b(ab)^na=a(ba)^m b =bea = ba. \]

Thus, since \((ba)^{n+1} = (ba)^n(ba)\) we can conclude that \((ba)^n(ba) = ba\) and then by the cancellation law, we have that \((ba)^n = e.\)

Now, to show that the order of \(ba\) is \(n\), we need to demonstrate that \(n\) is the smallest positive integer such that \((ba\)^n = e. )

Suppose there exists a positive integer \(m < n\) such that \((ba)^m = e.\) Then,\[ (ab)^{m+1} =(ab)(ab)\cdots(ab)=a\underbrace{(ba)\cdots(ba)}_{m \text{ times}}b=a(ba)^m b=aeb= ab. \]

Thus, since\((ab)^{m+1} =(ab)^m(ab)\) we have that \((ab)^m(ab)= ab\) and then by the cancellation law, we have that \((ab)^m=e\) which contradicts the fact that \(n\) is the smallest positive integer such that \((ab)^n=e.\)

Hence, \(n\) is the smallest positive integer such that \((ba)^n=e\) and therefore \((ba)\) has finite order \(n\).

Describe all groups G that contain no proper subgroup

Solution:

Prove that every subgroup of a cyclic group is cyclic. Do this by working with exponents, and use the description of the subgroups of \((\mathbb{Z},+)\).

1. Let \(G\) be a cyclic group of order 6. How many of its elements generate \(G\)?
   Answer the same question for cyclic groups of orders 5 and 8.
2. Describe the number of elements that generate a cyclic group of arbitrary order \(n\).

Solution:

b)

Let \(x\) and \(y\) be elements of a group \(G\). Assume that each of the elements \(x\), \(y\), and \(xy\) has order 2. Prove that the set \(H = \{I, x, y, xy\}\) is a subgroup of \(G\), and that it has order 4.

1. Prove that the elementary matrices of the first and third types (1.2.4) generate \(GL_n(\mathbb{R})\).
2. Prove that the elementary matrices of the first type generate \(SL_n(K)\). Do the \(2 \times 2\) case first.

Recall the types of elementary matrices,

How many elements of order \(2\) does the symmetric group \(S_4\) contain?

Show by example that the product of elementsof finite order in a group need not have finite order. What if the group is abelian?

Homomorphisms

Let \(\varphi: G \rightarrow G'\) be a surjective homomorphism. Prove that if \(G\) is cyclic, then \(G'\) is cyclic, and if \(G\) is abelian, then \(G'\) is abelian.

Prove that the intersection \(K \cap H\) ofsubgroupsof a group \(G\) is a subgroup of \(H\), and that if \(K\) is a normal subgroup of \(G\), then \(K \cap H\) is a normal subgroup of \(H\).

Let \(U\) denote the group of invertible upper triangular \(2 \times 2\) matrices \(A = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}\), and let \(\varphi : U \rightarrow \mathbb{R}^{\times}\) be the map that sends \(A\) to \(a^2\). Prove that \(\varphi\) is a homomorphism, and determine its kernel and image.

Let \(f: (\mathbb{R},+) \rightarrow (\mathbb{C},\times)\) be the map \(f(x) = e^{ix}\). Prove that \(f\) is a homomorphism, and determine its kernel and image.

Prove that the \(n \times n\) matrices that have the block form \(M = \begin{bmatrix} A & B \\ 0 & D \end{bmatrix}\), with \(A \in GL_r(\mathbb{R})\) and \(D \in GL_{n-r}(\mathbb{R})\), form a subgroup \(H\) of \(GL_n(\mathbb{R})\), and that the map \(H \rightarrow GL_r(\mathbb{R})\) that sends \(M \mapsto A\) is a homomorphism. What is its kernel?

Determine the center of \(GL_n (\mathbb{R})\).

Hint: You are asked to determine the invertible matrices \(A\) that commute with every invertible matrix \(B\). Do not test with a general matrix \(B\). Test with elementary matrices.

Isomorphisms

Let \(G'\) be the group of real matrices of the form \(\begin{bmatrix}1 & x \\ & 1\end{bmatrix}\) Is the map \(\phi: (\mathbb{R},+) \rightarrow G'\) that sends \(x\) to this matrix an isomorphism?

Describe all homomorphisms \(\phi : (\mathbb{Z},+) \rightarrow (\mathbb{Z},+)\). Determine which are injective, which are surjective, and which are isomorphisms.

Show that the functions \(f = \frac{1}{x}\), \(g = \frac{x - 1}{x}\) generate a group of functions, the law of composition being composition of functions, that is isomorphic to the symmetric group \(S_3\).

There are 6 elemenst in the \(S_3\). Let G be the group that define in the question. First find the all 6 elements group. Let’s do some rough works. \[\begin{eqnarray*} f(x)&=\frac{1}{x}\\\\ g(x)&=&\frac{x - 1}{x}\\\\ f\circ f(x)&=& \frac{1}{1/x}=x\\\\ g\circ g (x) &=& \frac{((x-1)/x)-1}{(x-1)/x}=\frac{1}{x-1}\\\\ f\circ g (x) &=& \frac{x}{x-1}\\\\ g\circ f (x) &=& \frac{(1/x)-1}{(1/x)}=1-x \end{eqnarray*}\] Now let’s define \[f_1=\frac{1}{x},~~f_2=\frac{x - 1}{x},~~ f_3=x,~~ f_4=\frac{1}{x-1}, ~~ f_5=\frac{x}{x-1}, ~~ , f_6=1-x\]. Now let’s construct multiplication table.

\(\circ\)	\(f_1\)	\(f_2\)	\(f_3\)	\(f_4\)	\(f_5\)	\(f_6\)
\(f_1\)	\(f_3\)	\(f_5\)	\(f_1\)	\(f_6\)	\(f_2\)	\(f_4\)
\(f_2\)	\(f_6\)	\(f_4\)	\(f_2\)	\(f_3\)	\(f_1\)	\(f_5\)
\(f_3\)	\(f_1\)	\(f_3\)	\(f_3\)	\(f_4\)	\(f_5\)	\(f_6\)
\(f_4\)	\(f_5\)	\(f_3\)	\(f_4\)	\(f_2\)	\(f_6\)	\(f_1\)
\(f_5\)	\(f_4\)	\(f_6\)	\(f_5\)	\(f_1\)	\(f_3\)	\(f_2\)
\(f_6\)	\(f_2\)	\(f_1\)	\(f_6\)	\(f_5\)	\(f_4\)	\(f_3\)

This table proves the closure property. According to the table \(f_3\) is the identity. Further, \[\begin{eqnarray*} \left(f_1\right)^{-1}&=&f_1\\ \left(f_2\right)^{-1}&=&f_4\\ \left(f_3\right)^{-1}&=&f_3\\ \left(f_4\right)^{-1}&=&f_2\\ \left(f_5\right)^{-1}&=&f_5\\ \left(f_6\right)^{-1}&=&f_6\\ \end{eqnarray*}\] The associativity property holds for the composition of rational functions are associative. To see the isomorphism let’s rearrange rows and columns in this table.

By looking at the table we can tell that \(G\cong S_3\). \(G\) is isomorphic to \(S_3\).

Let \(\phi:G\to S_3\) ide defined by \(\phi(f)=(23)\) and \(\phi(g)=(132)\)

Prove that in a group, the products \(ab\) and \(ba\) are conjugate elements.

Decide whether or not the two matrices \(A = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 1 \\ -2 & 4 \end{bmatrix}\) are conjugate elements of the general linear group \(GL_2(\mathbb{R})\).

Let \(H\) be a subgroup of \(G\), and let \(g\) be a fixed element of \(G\). The conjugate subgroup \(gHg^{-1}\) is defined to be the set of all conjugates \(ghg^{-1}\), with \(h\) in \(H\). Prove that \(gHg^{-1}\) is a subgroup of \(G\).

Prove that the map \(A \mapsto (A^{t})^{-1}\) is an automorphism of \(GL_{n}(\mathbb{R})\).

Prove that a group \(G\) and its opposite group \(G^{o}\) (Exercise 2.2.6) are isomorphic.

Find all automorphisms of

1. a cyclic group of order 10,
2. the symmetric group \(S_{3}\).

Let \(a\) be an element of a group \(G\). Prove that if the set \(\{1, a\}\) is a normal subgroup of \(G\), then \(a\) is in the center of \(G\).

Equivalence Relations and Partitions

Let \(G\) be a group. Prove that the relation \(a \sim b\) if \(b = gag^{-1}\) for some \(g\) in \(G\) is an equivalence relation on \(G\).

An equivalence relation on \(S\) is determined by the subset \(R\) of the set \(S \times S\) consisting of those pairs \((a, b)\) such that \(a \sim b\). Write the axioms for an equivalence relation in terms of the subset \(R\).

With the notation of Exercise 2.7.2, is the intersection \(R \cap R'\) of two equivalence relations \(R\) and \(R'\) an equivalence relation? Is the union?

A relation \(R\) on the set of real numbers can be thought of as a subset of the \((x, y)\)-plane. With the notation of Exercise 7.2, explain the geometric meaning of the reflexive and symmetric properties.

With the notation of Exercise 7.2, each of the following subsets \(R\) of the \((x, y)\)-plane defines a relation on the set \(\mathbb{R}\) of real numbers. Determine which of the axioms (2.7.3) are satisfied:

1. the set \(\{(s, s) | s \in \mathbb{R}\}\),
2. the empty set,
3. the locus \(\{xy + 1 = 0\}\),
4. the locus \(\{x^2y - xy^2 - x + y = 0\}\).

How many different equivalence relations can be defined on a set of five elements?

Cosets

Let \(H\) be the cyclic subgroup of the alternating group \(A_4\) generated by the permutation \((123)\). Exhibit the left and the right cosets of \(H\) explicitly.

First let’s see the elements in the \(A_4\). There are \(\frac{4!}{2}=\frac{1\times 2\times 3\times 4}{2}=\frac{24}{2}=12\) elements in the \(A_4\). \[ A_4 = \{id, (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243) \}\] let \(H=<(1 2 3)>=\{id, (1 2 3), (1 3 2) \}\) Let’s find the left cosets, we apply the elements of \(A_4\) to \(H\),

• \(id ~H = \{ id \cdot id, id \cdot (123), id \cdot (132)\} = \{ id, (123), (132) \}=H\)
• \((12)(34)H = \{ (12)(34) \cdot id, (12)(34) \cdot (123), (12)(34) \cdot (132)\}=\{(12)(34),(2 4 3),(1 4 3)\}\)
• \((13)(24)H = \{ (13)(24) \cdot id, (13)(24) \cdot (123), (13)(24) \cdot (132)\}= \{(13)(24),(142),(234)\}\)
• \((14)(23)H = \{ (14)(23) \cdot id, (14)(23) \cdot (123), (14)(23) \cdot (132)\}=\{(14)(23),(134),(1 2 4)\}\)
• \((123)H = \{ (123) \cdot id, (123) \cdot (123), (123) \cdot (132)\} =\{(123),(132),id\}=H\)
• \((132)H = \{ (132) \cdot id, (132) \cdot (123), (132) \cdot (132)\} =\{(132),id,(123)\}= H\)
• \((124)H = \{ (124) \cdot id, (124) \cdot (123), (124) \cdot (132)\}=\{(124),(14)(23),(134)\}\)
• \((142)H = \{ (142) \cdot id, (142) \cdot (123), (142) \cdot (132)\}=\{(142),(234),(13)(24)\}\)
• \((134)H = \{ (134) \cdot id, (134) \cdot (123), (134) \cdot (132)\}=\{(134),(124),(14)(23)\}\)
• \((143)H = \{ (143) \cdot id, (143) \cdot (123), (143) \cdot (132)\}=\{(143),(12)(34),(243)\}\)
• \((234)H = \{ (234) \cdot id, (234) \cdot (123), (234) \cdot (132)\}=\{(234),(13)(24),(142)\}\)
• \((243)H = \{ (243) \cdot id, (243) \cdot (123), (243) \cdot (132)\}=\{(243),(143),(12)(34)\}\)

As a summary, \[\begin{eqnarray*} id~H&=&(123)H=(132)H=\{id,(123),(132)\}=H\\ (12)(34)H&=&(243)H=(143)H=\{(12)(34),(243),(143)\}\\ (13)(24)H&=&(142)H=(234)H=\{(13)(24),(142),(234)\}\\ (14)(23)H&=&(134)H=(124)H=\{(14)(23),(134),(124)\} \end{eqnarray*}\] Futhur, \(A_4=H\sqcup(12)(34)H\sqcup(13)(24)H\sqcup(14)(23)H\). Similarly, \[\begin{eqnarray*} H~id&=&H(123)=H(132)=\{ id, (123), (132) \}=H\\ H(12)(34)&=& H (134)= H(234)=\{(12)(34),(134),(234)\}\\ H(13)(24)&=&H(243)=H(124)=\{(13)(24),(243),(124)\}\\ H(14)(23)&=&H(142)=H(143)=\{(14)(23),(142),(143)\} \end{eqnarray*}\] \(A_4=H\sqcup H(12)(34)\sqcup H (13)(24)\sqcup H(14)(23)\).\\ Further, note that \((12)(34)H\neq H(12)(34)\)

In the additive group \(\mathbb{R}^m\) of vectors, let \(W\) be the set of solutions of a system of homogeneous linear equations \(AX = 0\). Show that the set of solutions of an inhomogeneous system \(AX = B\) is either empty, or else it is an (additive) coset of \(W\).

Solution:
Let \(W=\{\underline{x}\in\mathbb{R}^n:A\underline{x}=0\}\) and \(W'=\{\underline{x}\in\mathbb{R}^n:A\underline{x}=B\}\)\ : Either \(W'=\emptyset\) or \(W'=\underline{v_0} +W\) for some \(\underline{v_0}\in \mathbb{R}^n\)\ Suppose that \(W'\neq \emptyset\).

Let \(\underline{x_1}\in W'\), then \(A\underline{x_1}=B\). Now consider following \[\underline{x_1}+W=\{\underline{x_1}+\omega:\omega\in W\}\] First of all, we are going to show that \(\underline{x_1}+W \subseteq W'\). Let \(\underline{x_1}+\underline{\omega_1}\in \underline{x_1}+W\). Now consider \[A(\underline{x_1}+\underline{\omega_1})=A(\underline{x_1})+A(\underline{\omega_1})=B+0=B\]. Thus, \(\underline{x_1}+\underline{\omega_1}\in W'\). Therefore, \(\underline{x_1}+W\subseteq W'\).\ Now we need to show that reverse inclusion. Let \(\underline{y_1}\in W'\). Now let \(\underline{\omega_2}=\underline{y_1}-\underline{x_1}\).Then \(A(\underline{\omega_2})=A(\underline{y_1}-\underline{x_1})=A(\underline{y_1})-A(\underline{x_1})=B-B=0\). Hence \(\underline{\omega_2}\in W\). \[ \underline{x_1}+\underline{\omega_2}=\underline{x_1}+(\underline{y_1}-\underline{x_1}) =y \in \underline{x_1}+W \] Hence, \(W'\subseteq \underline{x_1}+W\). Therefore, \(W'=\underline{x_1}+W\).

Does every group whose order is a power of a prime \(p\) contain an element of order \(p\)?

Yes.
Let’s prove following.
Claim :Every group whose order is a power of a prime \(p\) contain an element of order \(p\).

(Prooo of Claim) Let \(G\) be group with \(|G|=p^k<\infty\) for some \(k\in \mathbb{Z}^+\), where \(p\) is a prime number. Let \(e\neq x \in G\). Then by Artin’s book Corollary 2.8.10, \(Ord(x)|p^k\). Thus, \(Ord(x)\cdot q=p^k\) forsome \(q\in \mathbb{Z}\). Since \(p\) is a prime, \(Ord(x)=p^m\) for some \(1\leq m\leq k\). Choose \(y:=x^{(p^{m-1})}\). Now we are going to proove that \(Ord(y)=p\). \[y^p=\left(x^{(p^{m-1})}\right)^p=x^{(p^{m-1})\cdot p^1}=x^{p^m}=1\] If \(q\in \mathbb{Z}^+\) such that \(y^q=1\), then \[1=y^q=\left(x^{(p^{m-1})}\right)^q=x^{(q \cdot p^{(m-1)})}\] Since \(ord(x)=p^m\) then \[\begin{eqnarray*} q\cdot p^{(m-1)}&\geq & p^m \\ q &\geq & p \end{eqnarray*}\] Hence \(Ord(y)=p\). The group \(G\) whose order is a power of a prime \(p\) contain an element of order \(p\).

Does a group of order \(35\) contain an element of order \(5\)? of order \(7\)?

Claim: Let \(G\) be a group with order \(35\). \(G\) contains elements of order 5 & 7.

(proof of the claim.) Let \(x \in G\). By Artin’s book Corollary 2.8.10, \(Ord(x)\) should be a divisor of \(35\). In other words, \(Ord(x) \in \{1,5,7,35\}\).

• Case-I: Existence of group elements with order 5 We are going to use proof by contradiction method. Assume that there is no any element with order 5.
  • Sub claim: All non-identity elements have order 7. If \(e \neq p \in G\) with \(Ord(p) = 35\) then \(q = p^7\) has order 5 (\(\because q^5 = (p^{7})^5 = p^35 = e\)). This contradicts our hypothesis. Thus, in case-I all non-identity elements have order 7. Let \(h \in G\) with \(Ord(h) = 7\). Now consider cyclic subgroup generated from \(<h> = H\). Then \(|H| = 7\). Choose \(e \neq g \not \in H\). Then by above subclaim \(g\) has order 7. Now consider left cosets \(H, gH, g^2H, g^3H, g^4H, g^5H, g^6H\). Note that those cosets are disjoint. Because if cosets are disjoint then there exist elements such as, \(g^ah^n = g^bh^m\) for some \(1 \leq a, b \leq 6\) and \(1 \leq n, m \leq 6\). Then

\[\begin{eqnarray*} g^ah^n &=& g^bh^m \\ g^{a-b} &=& h^{m-n} \\ \end{eqnarray*}\]

So, we can choose \(r \in \mathbb{Z}^+\) such that \(r(a-b) \equiv 1 \pmod{7}\). Thus,

\[ g = g^{r(a-b)} = h^{r(m-n)} \in H \]

This contradicts selection of \(g\). Thus, \(H, gH, g^2H, g^3H, g^4H, g^5H, g^6H\) left cosets are disjoint. Recall the counting formula

\[\begin{eqnarray*} \text{(order of G)} &=& \text{(order of H)} \cdot \text{(number of cosets)} \\ |G| &=& |H| \cdot [G:H] \end{eqnarray*}\]

But observe that \(|G| = 35 \neq 49 = 7 \cdot 7 = |H| \cdot [G:H]\). This contradicts our first hypothesis. Thus \(G\) contains an element of order 5.

• Case-II: Existence of group elements with order 7 This case is also similar to previous one. But I am going to do it. Now assume that there is no any element with order 7. It is very similar to previous case we can prove that all non-identity elements have order 5. Let \(h \in G\) with order 5. We can choose \(g\) very similar to previous case. Further we can consider left cosets \(H, gH, g^2H, g^3H, g^4H\). So, we can prove that above cosets are disjoint. Then again we use counting formula, \(|G| = 35 \neq 25 = 5 \cdot 5 = |H| \cdot [G:H]\). This is a contradiction. Thus \(G\) contains an element of order 7.

A finite group contains an element \(x\) of order \(10\) and also an element \(y\) of order \(6\). What can be said about the order of \(G\)?

Let \(G\) b group, with \(|G|<\infty\). Let \(x,y\in G\) with \(|x|=10\) and \(|y|=6\). By Artin’s book Corollary 2.8.10 , \(|x|\biggr | |G|=10\biggr ||G|\) and \(|y|\biggr |G|=6\biggr | |G|\). Hence, \[LCM(|x|,|y|)\biggr |G|=LCM(10,6)\biggr | |G|=30\biggr | |G|\].

Let \(\phi: G \rightarrow G'\) be a group homomorphism. Suppose that \(|G| = 18\), \(|G'| = 15\), and that \(\phi\) is not the trivial homomorphism. What is the order of the kernel?

Then possibilities for \(Img(\phi)\) are as follows. We use Artin’s Algbera book Corollary 2.8.13, \[\begin{eqnarray*} |img(\phi)| \biggr ||G|~~~~&\text{and}&~~~~|img(\phi)| \biggr | |G'|\\ |img(\phi)| \biggr | 18~~~~&\text{and}&~~~~|img(\phi)| \biggr | 15\\ |img(\phi)| =1,2,3,6,9 \text{ or } 18~~~~&\text{and}&~~~~|img(\phi)| = 1,3,5\text{ or }15. \end{eqnarray*}\] Thus, \(|img(\phi)| =1,3\). But given that \(\phi\) is not the trivial homomorphism. Hence, \(|img(\phi)| \neq 1\). Therefore, \(|img(\phi)| =3\).
Now let’s find the order of the kernel. By Corollary 2.8.13, \[\begin{eqnarray*} |G | &=& |\ker(\phi)| \cdot |im(\phi)|\\ 18 &=& |\ker(\phi)| \cdot 3\\ |\ker(\phi)| &=& 6 \end{eqnarray*}\]

A group \(G\) of order 22 contains elements \(x\) and \(y\), where \(x \neq 1\) and \(y\) is not a power of \(x\). Prove that the subgroup generated by these elements is the whole group \(G\).

Let \(G\) of order 22 contains elements \(x\) and \(y\), where \(x \neq 1\) and \(y\) is not a power of \(x\). Let \(G'=<x,y>\).

Need to show: \(|H|=22\).
By Largrage Therom, \(|G'| \biggr | |G|=|G'| \biggr | 22\). So, \(|H|=1,2,11,22\). Let’s condier each and evrry cases.

• \(|H|\neq 1\). Beacuse \(1\neq x\in H\). So, \(H\) can not be happen trivial group.
• \(|H|\neq 2\). If \(|H|=2\) then \(y=1\) or \(y=x\).
  But both cases can not happen.So, \(|H|\neq 2\). Beacuse that contrduct that \(y\) is not a power of \(x\).
• \(|H|\neq 11\). Beacuse if \(|H|= 11\) and 11 is prime, then \(H\) is cyclic. (By Corollary 2.8.11) Furthur, \(H=<x>\). Then \(y\) is a power of \(x\), but it can not happen. So, \(|H|\neq 11\).
• Therefore, \(|H|=|<x,y>|=22\).

Let \(G\) be a group of order \(25\). Prove that \(G\) has at least one subgroup of order \(5\), and that if it contains only one subgroup of order \(5\), then it is a cyclic group.

Claim 1: Let \(G\) be a group of order 25. Then \(G\) has at least one subgroup of order 5.

Let \(G\) be a group of order 25. Let \(1 \neq x \in G\). By corollary 2.8.10, \(|x|\) divides \(|x| = |G'|\) which divides \(25\). So, \(|x| = 5, 25\). Let’s consider each and every case.

• If \(|x| = 5\) then \(H = <x>\). Then \(|H| = |<x>| = 5\) (By corollary 2.8.11), \(G\) has at least one subgroup of order 5.
• Suppose that \(|x| = 25\), let \(y = x^5\).

Subclaim: \(|y| = 5\).

\(y^5 = (x^5)^5 = x^{25} = 1\).

Let \(m \in \mathbb{Z}\) such that \(y^m = 1\). Then \(y^m = (x^5)^m = x^{5m}\). Then \(25 \geq 5m \implies 5 \geq m\). Thus, \(|y| = 5\).

Hence \(<y> = <x^5>\) is a subgroup of order 5.

Claim 2 :If it contains only one subgroup of order 5, then it is a cyclic group. ::: {.proof} Suppose that \(G\) contains only one subgroup of order 5. Let’s call that subgroup \(G'\). Let \(y \not\in G'\) and \(1 \neq y \in G\). By corollary 2.8.10, \(|y|\) divides \(|G| = |y|\) which divides \(25\). So, \(|y| = 5, 25\).

• If \(|y| = 5\), then \(|<y>| = 5\). But \(G'\) is the only one subgroup of order 5. \(G' = <y>\). But it contradicts the selection of \(y\).
• If \(|y| = 25\), then \(|<y>| = 25\), so \(<y> = G\). Therefore, \(G = <x>\) is cyclic.
  :::

Let \(G\) be a finite group. Under what circumstances is the map \(f: G \to G\) defined by \(f(x) = x^2\) an automorphism of \(G\)?

• Claim 1: The map \(f:G\to G\) is a homomorphism if and only if G is abelian.

\(f\) is a homomorphism if and only if for all \(x, y \in G\), \(f(xy) = f(x)f(y)\). This is equivalent to for all \(x, y \in G\), \((xy)^2 = (xy)(xy) = x^2y^2\). This is equivalent to for all \(x, y \in G\), \(xyxy = xxyy\). This is equivalent to for all \(x, y \in G\), \(xy = yx\). This is equivalent to \(G\) is abelian.

• Claim 2: \(f\) is injective if and only if no element in \(G\) has order \(2\). ::: {.proof} \(f\) is injective if and only if \(\ker(f) \neq \{1\}\). This is equivalent to for all \(x \in G\) with \(1 \neq x\), \(f(x) = x^2 \neq 1\). This is equivalent to no element having order 2. This is equivalent to all elements of \(G\) having odd order (By following sub claim).

  • Sub Claim 2.1: Group \(H\) has no element having order \(2\) is if and only if all elements of H having odd order.

First let’s prove the backward direction. We use proof of contradiction. Assume the contrary. There are some elements of \(H\) having even order. So, \(|x| = 2k\) for some \(k \in \mathbb{Z}^+\). Then \(|x^k| = 2\) has order \(2\). :::

Prove that every subgroup of index \(2\) is a normal subgroup, and show by example that a subgroup of index \(3\) need not be normal.

Let \(G\) be a group and let \(H \leq G\) with \([G:H] = 2\). So, \(H\) has only two left cosets. Then left cosets are \(H\) and \(G \setminus H\) and the cosets partition \(G\). Let \(x \in G\).

• If \(x \in H\), then \(gH = H = Hg\).
• If \(x \notin H\), then \(gH = G \setminus H\) and \(Hg = G \setminus H\). Thus, \(gH = Hg\).

Therefore, \(G\) is a normal subgroup.

Claim: A subgroup of index 3 need not be normal.

Example: Consider \(S_3 = \{id, (12), (13), (23), (123), (132)\}\) and \(H = <(12)> = \{id, (12)\} \leq S_3\). Let’s find the index using the counting formula

\[\begin{align*} |G| &= |H| \cdot [G:H] \\ 6 &= 2 \cdot [G:H] \\ [G:H] &= 3 \end{align*}\]

Let’s take \(g = (13)\) and \(h = (12)\), then \(ghg^{-1} = (13)(12)(13) = (23)\). But \((23) \notin H\), so \(H\) is not a normal subgroup of \(S_3\). Therefore, we have proven that \(H\) is not a normal subgroup of \(S_3\). Thus, \(H\) is not a normal subgroup.

Let \(G\) and \(H\) be the following subgroups of \(GL_2 (\mathbb{R})\): \[ G =\left\{ \begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix}\right\}, H = \left\{\begin{bmatrix} x & 0 \\ 0 & 1 \end{bmatrix}\right\} \] with \(x\) and \(y\) real and \(x > 0\). An element of \(G\) can be represented by a point in the right half plane. Make sketches showing the partitions of the half plane into left cosets and into right cosets of \(H\).

Let \(S\) be a subset of a group \(G\) that contains the identity element 1, and such that the left cosets \(aS\), with \(a\) in \(G\), partition \(G\). Prove that \(S\) is a subgroup of \(G\).

Let \(1 \in S \subseteq G\), and such that \(\Pi = \{aS : a \in G\}\), is a partition of \(G\). -Subset : This is trivial.(Given in the problem in already.) - Identity : \(1 \in S\). - Closure and Inversion : Let \(x, y \in S\). Since \(\Pi\) is part of \(G\), there exists \(g \in G\) such that \(xy^{-1} \in gS\). So, \(xy^{-1} = gr\) for some \(r \in S\). Then \(x = gry \in grS\) (since \(y \in S\)). Thus, \(x \in S \cap grS\). This implies \(S \cap grS \neq \emptyset\). But \(\Pi\) is a partition of \(G\). Thus, \(S = grS\). Now observe that \(gr \in gS\) (since \(r \in S\)) and \(gr \in grS\). Hence \(gr \in gS \cap grS\). Thus, \(gS = grS\). Hence, \(xy^{-1} = S = gS = grS\).

By the subgroup test, \(S\) is a subgroup of \(G\).

Let \(S\) be a set with a law of composition: A partition \(\Pi_1 \cup \Pi_2 \cup \cdots\) of \(S\) is compatible with the law of composition if for all \(i\) and \(j\), the product \[\Pi_i \Pi_j =\{ xy: x\in \Pi_i , y\in \Pi_j\}\] set is contained in a single subset \(\Pi_k\) of the partition.

1. The set \(Z\) of integers can be partitioned into the three sets [Pos] , [Neg], [{0}]. Discuss the extent to which the laws of composition \(+\) and \(\times\) are compatible with this partition.
2. Describe all partitions of the integers that are compatible with the operation \(+\)

Modular Arthmetic

What are the possible values of \(a^2\) modulo 4? modulo 8?

solution:

• In modulo 4
  \[\begin{eqnarray*} \overline{0}^2&\equiv & 0 \pmod{4}\\ \overline{1}^2&\equiv & 1 \pmod{4}\\ \overline{2}^2&\equiv & 0 \pmod{4}\\ \overline{3}^2&\equiv & 1 \pmod{4}\\ \end{eqnarray*}\] The possible values of \(a^2\pmod{4}\) are 0 and 1.

• In modulo 8
  \[\begin{eqnarray*} \overline{0}^2&\equiv & 0 \pmod{8}\\ \overline{1}^2&\equiv & 1 \pmod{8}\\ \overline{2}^2&\equiv & 4 \pmod{8}\\ \overline{3}^2&\equiv & 1 \pmod{8}\\ \overline{4}^2&\equiv & 0 \pmod{8}\\ \overline{5}^2&\equiv & 1 \pmod{8}\\ \overline{6}^2&\equiv & 4 \pmod{8}\\ \overline{7}^2&\equiv & 1 \pmod{8}\\ \end{eqnarray*}\] The possible values of \(a^2\pmod{8}\) are 0,1 and 4.

Prove that every integer \(a\) is congruent to the sum of its decimal digits modulo 9.

Let \(x\in \mathbb{Z}\). Now we can represent \(x\) as follows

\[x=a_010^0+a_110^1+\cdots +a_n10^n=\sum_{i=0}^n a_{i}10^i\text{ forsome } n\in \mathbb{Z},\text{ and }a_i\in \{0,1,...,9\}\]

We need to show \(x \equiv \sum_{i=0}^na_i\pmod{9}\). So, now consider,

\[\begin{eqnarray} x-\sum_{i=0}^n a_{i}&=&\sum_{i=0}^n a_{i}10^i-\sum_{i=0}^n a_{i}\\ &=& \sum_{i=0}^n (a_{i}10^i- a_{i})\\ &=& \sum_{i=0}^n a_{i}(10^i- 1) \end{eqnarray}\]

By follwing calim we can get that,

\[\begin{eqnarray} x-\sum_{i=0}^n a_i &\equiv & 0 \pmod{9}\\ x &\equiv & \sum_{i=0}^na_i\pmod{9} \end{eqnarray}\]

: \(9|(10^k-1)\) for any \(k\in \mathbb{N}\).

We use mathematical induction.
\(k=1\)
This case is trivial. Because \(9|10-1\).
\(k=n\in\mathbb{Z}\)
Now asuume when \(n=k, 9|(10^n-1)\).
\(k=n+1\) \[10^{n+1}-1 =10\cdot 10^{n}-1=9\cdot 10^n + (10^n - 1) \] Thus \(9\vert (10^{n+1}-1)\).
Therefore,by mathematical induction, \(9|(10^n-1)\) for any \(n\in \mathbb{N}\).

Solve the congruence \(2x \equiv 5\) modulo 9 and modulo 6.

Determine the integers \(n\) for which the pair of congruences \(2x - y \equiv 1\) and \(4x + 3y \equiv 2\) modulo \(n\) has a solution.

Prove the Chinese Remainder Theorem: Let \(a, b, u, v\) be integers, and assume that the greatest common divisor of \(a\) and \(b\) is 1. Then there is an integer \(x\) such that \(x \equiv u\) modulo \(a\) and \(X \equiv v\) modulo \(b\).
Hint: Do the case \(u = 0\) and \(v = 1\) first.

Determine the order of each of the matrices \(A=\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}\) and \(B=\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}\) when the matrix entries are interpreted modulo 3.

The Correspondence Theorem

Describe how to tell from the cycle decomposition whether a permutation is odd or even.

Let H and K be subgroups of a group G.

a)Prove that the intersection xH n yK of two cosets of H and K is either empty or else is a coset of the subgroup H n K. b) Prove that if H and K have finite index in G then H n K also has finite index in G.

Let \(G\) and \(G'\) be cyclic groups of orders \(12\) and \(6\), generated by elements \(x\) and \(y\), respectively, and let \(\phi : G \to G'\) be the map defined by \(\phi(x^i) = y^i\). Exhibit the correspondence referred to in the Correspondence Theorem explicitly.

With the notation of the Correspondence Theorem, let \(H\) and \(H'\) be corresponding subgroups. Prove that \[[G : H] = [G' : H']\].

With reference to the homomorphism S4 -> S3 described in Example 2.5.13, determine the six subgroups of S4 that contain K.

Product Group

Let \(x\) be an element of order \(r\) of a group \(G\), and let \(y\) be an element of \(G'\) of order \(s\). What is the order of \((x, y)\) in the product group \(G \times G'\)?

Solution: The order of \((x, y)\) in the product group \(G \times G'\) is \(lcm(r,s)\)
Let \(n\in \mathbb{Z}^+\) such that \[(x,y)^n=(x^n,y^n)=(1_G,1_{G'}).\] This implies, \(x^n=1_G\) and \(y^n=1_{G'}\). Since order of \(x\) and \(y\) are \(r\) and \(s\) respectively, \[r|n~~~\text{ and }~~~s|n\] So, we know that the least positive integer such that above property holds is \(lcm(r,s)\). Hence,The order of \((x, y)\) in the product group \(G \times G'\) is \(lcm(r,s)\).

What does Proposition @ref(prp:prp2114) tell us when, with the usual notation for the symmetric group \(S_3,K\) and \(H\) are the subgroups \(<y>\) and \(<x>\)?

Recall: \(y^2=1\) and \(x^3=1\). Then \[H=<x>=\{1,x,x^2\}~~~\text{ and }~~~ K=<y>=\{1,y\}\] Let the multiplication map, \(f:H\times K \to S_3\) defined by \(f(h,k)=hk\). Then, \(Im(f)=HK=\{hk:h\in <x>, k\in <y>\}=\{1,x,x^2,y,xy,x^2y\}\).

Claim 1: \(f\) is injective.
Observe that \(H \cap K=\{1\}\). by @ref(prp:prp2114) a) \(f\) is injective.

Claim 2: \(f\) is surjective.
Observe that \(S_3=HK=\{1,x,x^2,y,xy,x^2y\}\). Thus, \(f\) is surjective.

Claim 3 : \(f\) is not homomorphism.
\[\begin{eqnarray} f((x,y)\cdot (x,1))&=& f(x^2,y^2)=f(x^2,1)=x^2\cdot 1=x^2\\ f(x,y)\cdot f(x,1)&=& (x\cdot y)\cdot (x\cdot 1)=(xy)\cdot x=y \end{eqnarray}\] Thus, \(f\) is not a homomorphism.

Prove that the product of two infinite cyclic groups is not infinite cyclic.

• Claim 1 : Infinite cyclic groups are isomorphic to \(Z\).
  Let \(G\) be a infinte cyclic group and \(<a>=G\). We can define a function \(f: \mathbb{Z} \rightarrow G\) by \(f(n) = a^n\) for all integers \(n\). We need to show that \(f\) is an isomorphism, which means it is bijective.

  • Subclaim 1.1 : \(f\) is a hormorphism.
    \[f(n+m)=a^{n+m}=a^n \cdot a^m =f(n)\cdot f(m)\] Thus, \(f\) is a hormorphism.
  • Subclaim 1.2 :\(f\) is a injective
    Suppose that \(f(n)=f(m)\) \[\begin{eqnarray} f(n)&=&f(m)\\ a^n &=& a^m \\ n&=& m ~~(\text{Since order is infinite}) \end{eqnarray}\]
  • Subclaim 1.3 : \(f\) is surjective.
    Let \(x\in G=<a>\). Then, \(x=a^k\) for some \(k\in \mathbb{Z}\). Then, observe that \[f(k)=a^k=a \]. Thus, \(f\) is surjective.

Therefore, \(f\) is an isomorphism. Thus, the infinite cyclic groups are isomorphic to \(\mathbb{Z}\).

We can consider \(\mathbb{Z}\times \mathbb{Z}\). Suppose that \((a, b)\) generate the product group. But then, we see that \((2a, b)\) cannot be obtained from adding \((a, b)\) to itself, which implies that \(\mathbb{Z}\times \mathbb{Z}\) is not infinite cyclic.

In each of the following cases, determine whether or not \(G\) is isomorphic to the product group \(H \times K\).

1. \(G = \mathbb{R}^\times\), \(H = \{\pm 1\}\), \(K = \{\)positive real numbers\(\}\).

2. \(G = \{\)invertible upper triangular \(2 \times 2\) matrices\(\}\), \(H = \{\)invertible diagonal matrices\(\}\), \(K = \{\)upper triangular matrices with diagonal entries \(1\}\).

3. \(G = e^x\), \(H = \{\)unit circle\(\}\), \(K = \{\)positive real numbers\(\}\).

Solution:

1. Observe that \(G=\mathbb{R}^\times=\mathbb{R}\setminus\{0\}\) is abelian. Then \(H,K \trianglelefteq G\). Further, observe that \(H\cap K=\{1\}\) and \(HK=\mathbb{R}^\times=G\). Thus, by proposition @ref(prp:prp2114) \(G \cong H \times K\).

• Claim b.1: \(K\trianglelefteq G\).
  Let \(g = \begin{bmatrix} a & b \\ & c \end{bmatrix}\) be an arbitary matrix in \(G\) and \(k = \begin{bmatrix} 1 & d \\ & 1 \end{bmatrix}\in K\). Then, \[ g^{-1} = \begin{bmatrix} a^{-1} & -ba^{-1}c^{-1} \\ & c^{-1} \end{bmatrix}\] Then \[gkg^{-1} = \begin{bmatrix} a & b \\ & c \end{bmatrix}\begin{bmatrix} 1 & d \\ & 1 \end{bmatrix}\begin{bmatrix} a^{-1} & -ba^{-1}c^{-1} \\ & c^{-1} \end{bmatrix}\] \[= \begin{bmatrix} a & b \\ & c \end{bmatrix}\begin{bmatrix} a^{-1} & dc^{-1} - ba^{-1}c^{-1} \\ & c^{-1} \end{bmatrix} = \begin{bmatrix} 1 & adc^{-1} \\ & 1 \end{bmatrix} \in K\] Thus, \(H\trianglelefteq G\).

• Claim b.2 : \(H=Z(G)\) (\(H\) is in the center of \(G\)).

  • subclaim b.2.1 : \(H \subseteq Z(G)\).
    Let \(h=\begin{bmatrix} x & 0 \\ & x \end{bmatrix} \in H\). Then, \[B=\begin{bmatrix} x & 0 \\ & x \end{bmatrix}=x\begin{bmatrix} 1 & \\ & 1 \end{bmatrix}=BI\] Then observe that, for any \(A\in G\), \[BA=(xI)A = xA = Ax = A(xI)=AB\]
  • subclaim b.2.2 : \(H \supseteq Z(G)\).
    Let \(\begin{bmatrix} p & q \\ r & s \end{bmatrix}\in Z(G)\), then,

\[\begin{eqnarray} \begin{bmatrix} p & q \\ r & s \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} &=& \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} p & q \\ r & s \end{bmatrix}\\ \begin{bmatrix} p & -q \\ r & -s \end{bmatrix}&=& \begin{bmatrix} p & q \\ -r & -s \end{bmatrix}\\ \end{eqnarray}\] This yields \(-q = q\) and \(r = -r\) which imply \(q = r = 0\). Now considering,

\[\begin{eqnarray} \begin{bmatrix} p & 0 \\ 0 & s \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} &=& \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} p & 0 \\ 0 & s \end{bmatrix}\\ \begin{bmatrix} 0 & p \\ q & 0 \end{bmatrix}&=& \begin{bmatrix} 0 & q \\ p & 0 \end{bmatrix}\\ \end{eqnarray}\] This implies that \(a = d\) and since the matrix is invertible, \(a\neq 0\). Thus, \(H \supseteq Z(G)\).

Therefore, \(H = Z(G)\). In other words, \(H\) is in the center of \(G\).

• Claim b.2 : \(H\trianglelefteq G\).
  Since, \(H\) is in the center of \(G\), then \(H\) is normal.

• Claim b.3 : \(H\cap K=\{I\}\).
  This is trivial.

• Claim b.4 : \(HK=G\) Clearly, \(HK\subseteq G\). Conversly, for any \(g\in G\), \(a\neq 0, b\neq 0\), \[g=\begin{bmatrix} a & b \\ & c \end{bmatrix}=\begin{bmatrix} a & \\ & c \end{bmatrix}\begin{bmatrix} 1 & a^{-1}b \\ & 1 \end{bmatrix}\in HK\]. Thus, \(HK=G\).

Therefore, by proposition @ref(prp:prp2114) \(G\cong HK\)

3. Since \(C^\times\) is abelian, then \(H\) and \(K\) are normal. And, \(H \cap K = \left\lbrace 1 \right\rbrace\).

Consider \(a + bi \in C^\times\). Then \[a + bi = \left(\frac{a}{\sqrt{a^2+b^2}} + \frac{b}{\sqrt{a^2+b^2}}i\right)\sqrt{a^2+b^2} \in HK\] So, \(HK = G\). So, by proposition @ref(prp:prp2114) \(G \cong H \times K\).

Let \(G_1\) and \(G_2\) be groups, and let \(Z_i\) be the center of \(G_i\). Prove that the center of the product group \(G_1 \times G_2\) is \(Z_1 \times Z_2\).

Let \(G\) be a group that contains normal subgroups of orders \(3\) and \(5\), respectively. Prove that \(G\) contains an element of order \(15\).

Let \(H\) be a subgroup of a group \(G\), let \(\phi:G \to H\) be a homomorphism whose restriction to \(H\) is the identity map, and let \(N\) be its kernel. What can one say about the product map \(H \times N \to G\)?

Let \(G\), \(G'\), and \(H\) be groups. Establish a bijective correspondence between homomorphisms \(\phi : H \to G \times G'\) from \(H\) to the product group and pairs \((\varphi, \varphi')\) consisting of a homomorphism \(\varphi:H \to G\) and a homomorphism \(\varphi':H \to G'\).

Let \(H\) and \(K\) be subgroups of a group \(G\). Prove that the product set \(HK\) is a subgroup of \(G\) if and only if \(HK = KH\).

Solution :

Let \(G\) be group and \(H,K\leq G\).

• \((\Longrightarrow)\): Claim : if \(HK\leq G\) then \(HK=KH\).
  

  Suppose \(HK\leq G\).

  • sub claim: Let \(x\in KH\). Then \(x=kh\) for some \(h\in H\) and \(k\in K\), \[x=kh=1kh1=(1k)(h1)\in HK\] (Since \(1\in H\) and \(1\in K\)). Thus, \(KH \subseteq HK\).
    
  • sub claim: \(KH \supseteq HK\).
    We can not use previous method.Because we still do not know \(KH\) is sub group or not.
    Let \(y\in HK\). Since \(HK\leq G\), \(y^{-1}\in HK\). Then \(y^{-1}=hk\) for some \(h_0\in H\) and \(k_0\in K\). \ \[y=(y^{-1})^{-1}=(h_0k_0)^{-1}=k_0h_0\in KH\] Thus, \(KH \supseteq HK\).

Therefore, \(KH=HK\).

• \((\Longleftarrow)\): Claim : If \(HK=KH\) then \(HK\leq G\).
  

Now suppose that \(HK=KH\).Suppose that \(HK=KH\) is empty. Then, \(H,K\) is empty. Now we use sub group test.
- Closure: Let \(x,y\in HK\). Then \(x=h_1k_1\) and \(y=h_2k_2\) for some \(h_1,h_2\in H\) and \(k_1,k_2\in K\). \[\begin{eqnarray} xy&=&(h_1k_1)(h_2k_2)\\ &=& h_1(k_1h_2)k_2) ~~(\text{By associavity property})\\ &=& h_1(h_3k_3)k_2) ~\text{for some }h_3\in H \text{ and } k_3\in K ~(\text{Since} KH=HK)\\ &=& (h_1h_3)(k_3k_2)\in HK \end{eqnarray}\] - Inverse: Let Let \(x\in HK\). Then \(x=h'k'\) for some \(h'\in H\) and \(k'\in K\). \[x^{-1}=(h'k')^{-1}=k'h'\in KH=HK\] Thus, \(HK\leq G\).

Therefore, the product set \(HK\) is a subgroup of \(G\) if and only if \(HK = KH\)

Quotient Groups

Show that if a subgroup \(H\) of a group \(G\) is not normal, there are left cosets \(aH\) and \(bH\) whose product is not a coset.

Suppose that \(H \leq G\) but \(H \not\trianglelefteq G\). We are going to use indirect proof. So, assume that \((aH)(bH)\) is a coset for all \(a,b\in G\). Then, \[aHbH=abH\] then for all \(h_1,h_2,h_3\in H\), So, \[ah_1bh_2=abh_3\] \[\begin{eqnarray} ah_1bh_2 &= & abh_3\\ a^{-1}(ah_1bh_2) &=& a^{-1}(abh_3)\\ (a^{-1}a)(h_1bh_2) &=& (a^{-1}a)(bh_3)\\ h_1bh_2 &=& bh_3\\ b^{-1}(h_1bh_2)h_2^{-1} &=& b^{-1}(bh_3)h_2^{-1}\\ b^{-1}h_1b(h_2h_2^{-1}) &=& (b^{-1}b)h_3h_2^{-1}\\ b^{-1}h_1b &=& h_3h_2^{-1}\in H\\ \end{eqnarray}\] Thus, \(b^{-1}h_1b\in H\). and since \(b\in G\) and \(h_1\in H\) is arbitary, \(H\) is normal subgroup, which contradict the our hypothesis. Thus, there exist left cosets \(aH\) and \(bH\) whose product is not a coset.

In the general linear group \(GL_3(\mathbb{R})\), consider the subsets:

\[ H = \left\{ \begin{pmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{pmatrix} \mid * \text{ represents an arbitrary real number} \right\} \]

and

\[ K = \left\{ \begin{pmatrix} 1 & 0 & * \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \mid * \text{ represents an arbitrary real number} \right\} \]

Show that \(H\) is a subgroup of \(GL_3\), that \(K\) is a normal subgroup of \(H\), and identify the quotient group \(H/K\). Determine the center of \(H\).

• Claim 1: \(H \leq G\).
  Certainly! Let’s prove that the subset \(H\) is a subgroup of the general linear group \(GL_3(\mathbb{R})\).

  • Non-emptiness: We start by showing that \(H\) is non-empty. The identity matrix \(\mathbf{I}_3\) belongs to \(H\) since it satisfies the conditions for \(H\): \[ \mathbf{I}_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] Therefore, \(H\) is non-empty.

  • Closure under matrix multiplication: Let \(\mathbf{A}\) and \(\mathbf{B}\) be matrices in \(H\). We need to show that their product \(\mathbf{AB}\) is also in \(H\). Consider: \[ \mathbf{A} = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \quad \text{and} \quad \mathbf{B} = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \] The product \(\mathbf{AB}\) is: \[ \mathbf{AB} = \begin{pmatrix} 1 & a+x & by+az+b \\ 0 & 1 & c+z \\ 0 & 0 & 1 \end{pmatrix} \] ````` Since \(a+x\), \(y+az+b\), and \(c+z\) are arbitrary real numbers, \(\mathbf{AB}\) satisfies the conditions for \(H\). Hence, \(H\) is closed under matrix multiplication.

  • Closure under taking inverses: Let \(\mathbf{A}\) be a matrix in \(H\). We need to show that its inverse \(\mathbf{A}^{-1}\) is also in \(H\). The inverse of \(\mathbf{A}\) is:

\[ \mathbf{A}^{-1} = \begin{pmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{pmatrix} \] Again, since \(-a\), \(ac-b\), and \(-c\) are arbitrary real numbers, \(\mathbf{A}^{-1}\) satisfies the conditions for \(H\). Therefore, \(H\) is closed under taking inverses. Hence, we have shown that \(H\) is a subgroup of \(GL_3(\mathbb{R})\).

• Claim 2 : \(K\leq H\).
  I am not going to prove that this case, because this is exhausting. This is very easy.

• Claim 3 : \(K\) is normal sub group of \(H\).
  Let \(A:=\begin{bmatrix} 1 & a & b \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix}\in H\) and \(B:=\begin{bmatrix} 1 & 0 & d \\0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in K\), where \(a,b,c,d\in \mathbb{R}\). Then,

\[\begin{eqnarray} ABA^{-1}& =& \begin{bmatrix} 1 & a & b \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & d \\0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{bmatrix} \\ &=&\begin{bmatrix} 1 & a & d+b \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{bmatrix} \\ &=& \begin{bmatrix} 1 & 0 & d \\0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\in K \end{eqnarray}\] Thus, \(K\) is normal sub group of \(H\).

• Quotient group \(H/K\) :
  Let \(h_1,h_2\in H\) be such that \[h_1=\begin{bmatrix} 1 & a_1 & b_1 \\0 & 1 & c_1\\ 0 & 0 & 1\end{bmatrix} \text{ and } h_2=\begin{bmatrix} 1 & a_2 & b_2 \\0 & 1 & c_2\\ 0 & 0 & 1\end{bmatrix}.\] Suppose that \(h_1K=h_2K\). By Artin’s book 2.85, \(h_1^{-1}h_2\in K\). \[h_1^{-1}h_2= \begin{bmatrix} 1 & -a_1 & a_1c_1-b_1 \\0 & 1 & -c_1\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & a_2 & b_2 \\0 & 1 & c_2\\ 0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & -a_1+a_2 & b_2-a_1c_2+a_1c_1-b_1 \\0 & 1 & -c_1+c_2\\ 0 & 0 & 1\end{bmatrix}\] Since,

\[\begin{eqnarray} h_1K=h_2K & \iff & h_1^{-1}h_2\in K\\ & \iff & -a_1+a_2=0 \text{ and }-c_2-c_1\\ & \iff & a_1=a_2\text{ and }c_1=c_2 \end{eqnarray}\]

Thus, \[H/K=\left\{hK\Biggr | \begin{bmatrix} 1 & a & 0 \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix}, a,c\in \mathbb{R}\right\}\]

• Center of Group H
  \[Z(H)=\left\{A\in H\biggr | AX=XA \text{ for all } X\in H\right\}\] Let’s find center. We need to find \(a,b,c\in \mathbb{R}\) such that \[\begin{eqnarray} \begin{bmatrix} 1 & a & b \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & x & y \\0 & 1 & z\\ 0 & 0 & 1\end{bmatrix} &=& \begin{bmatrix} 1 & x & y \\0 & 1 & z\\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & a & b \\0 & 1 & c\\ 0 & 0 & 1\end{bmatrix} \text{ for all } x,y,z\in \mathbb{R}\\ \begin{bmatrix} 1 & a+x & b+y+az \\0 & 1 & c+z\\ 0 & 0 & 1\end{bmatrix} &=& \begin{bmatrix} 1 & a+x & b+y+xc \\0 & 1 & c+z\\ 0 & 0 & 1\end{bmatrix} \end{eqnarray}\] By comparing \(1 \times 3\) index, we need to find that \(a,b,c\in \mathbb{R}\) \(az=xc\) for all \(x,y,z\mathbb{R}\). SO, the only possibility is \(a=c=0\). Thus, \[Z(H)=\left\{\begin{bmatrix} 1 & 0 & b \\0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\Biggr | b\in \mathbb{R}\right\}\]

Let \(P\) be a partition of a group \(G\) with the property that for any pair of elements \(A\), \(B\) of the partition, the product set \(AB\) is contained entirely within another element \(C\) of the partition. Let \(N\) be the element of \(P\) that contains 1. Prove that \(N\) is a normal subgroup of \(G\) and that \(P\) is the set of its cosets.

• Claim 1: The product set \(NN=N\).
  Let \(x\in N\), Then \(x=x\cdot 1 \in NN\). Thus, \(N \subseteq NN\). We know that the product set \(NN\) is contained entirely within in an element of pratition, but \(N\) is a partion. The only way this happen \(N=NN\).

• Claim 2: \(N \leq G\).
  

  • Subset: It is very clear that \(N \subseteq G\).
  • Closure: Let \(x,y\in N\). Then \((xy)\in NN=N\). So, \(N\) is closed under composition.
  • Identity: Given that \(1\in N\).
  • Inverse: Let \(x\in N\). We have to show that \(x^{-1}\) is contained in \(N\). We use proof by contardiction. So, Assume contary, \(x^{-1}\in M\), \(M\) is an elemenet of partion \(P\) such that \(N\neq M\).

\[xx^{-1}\in NM, xx^{-1}=1 \in N\] So, we would have \(NM \subseteq N\) (since the product set is contained entirely wiithin one of the sets of partition). On the other hand, \[1x^{-1}\in NM, 1x^{-1}=x^{-1} \in N\] So, \(NM\subseteq N\)a contradiction.

Therefore, conclude that our assumption that \(x^{-1}\not\in N\) was wrong, that we must have \(x^{-1}\in N\)

• Claim 3: \(N\trianglelefteq\)
  Let \(a\in G\) and \(n\in N\). Let \(a\in A\) and \(a^{-1}\in B\), where \(A,B\in P\). \[ana^{-1}=ANB\], But also, \(1\in N\), \[g\cdot 1 \cdot g ^{-1}=1\in ANB\] Since P spilts into disjoint subsets and the only one element of partion that containn \(1\) is \(N\). \[ANB=N\] and but also \(gng^{-1}\in N\). Thefeore, \(N\) is normal sub group.

-Claim 4: P is the set of its cosets.
Let \(A\) be some element Of partition, and let \(a \in A\).

Need to prove: \(A=aN\)

• Subclaim 4.1: \(A\subseteq aN\).
  Let \(a\in A\). Let \(b\in B\), where \(B\) is elmenet of partion. Then \(b^{-l}b\in BA, \quad b^{-l} = 1 \in N\) So, \(BA \subseteq N\). Specially, \[b^{-1}a \in N\] So there exists some \(n \in N\) such that \[b^{-1}a = n \quad \iff \quad a =bn \quad \iff \quad b=an^{-1}\] Since \(N\leq G\). \(n^{-1}\in N\). Thus, \(b \in aN\), so \(A \subset aN\).

• Subcliam 4.2 : \(A\supseteq aN\).
  

  • subclaim 4.2.1: \(AN \subseteq A\),

Note that \(a \in A\) and \(1 \in N\), Thus, \[a\cdot 1\in AN, \quad a\cdot 1=a\in A\] Since, the product set is contained entirely in one element partition. Thus, \(AN\subseteq A\)

Now, \[aN=\{an | n \in N\} \subseteq \{an : a\in A, n\in N\} = AN \subseteq A \] Thus, \(ax \in A\), as required. Thus \(aN\subseteq A\)

Hence, we can conclude that \(A =aN\).

Let \(H = \{\pm1, \pm i\}\) be the subgroup of \(G = C^{\times}\) of fourth roots of unity. Describe the cosets of \(H\) in \(G\) explicitly. Is \(G / H\) isomorphic to \(G\)?

Let \(G\) be the group of upper triangular real matrices \(\begin{bmatrix} a & b \\ 0 & d \end{bmatrix}\) with \(a\) and \(d\) different from zero. For each of the following subsets, determine whether or not \(S\) is a subgroup, and whether or not \(S\) is a normal subgroup. If \(S\) is a normal subgroup, identify the quotient group \(G /S\).

1. \(S\) is the subset defined by \(b = 0\).
2. \(S\) is the subset defined by \(d = 1\).
3. \(S\) is the subset defined by \(a = d\).

Miscellaneous Problems

Describe the column vectors \((a, c)^T\) that occur as the first column of an integer matrix \(A\) whose inverse is also an integer matrix.

1. Prove that every group of even order contains an element of order \(2\).
2. Prove that every group of order \(21\) contains an element of order \(3\).

Claim a : Let \(G\) be a group whose identity is \(e\).Let \(G\) be of even order.Then,\(\exists x \in G: |x| = 2\)

(Claim a) In any group \(G\), the identity element \(e\) is self-inverse with the property that the identity is the only group element of order 1, and is the only such element.

That leaves an odd number of elements.

Each element in \(x \in G: |x| > 2\) can be paired off with its inverse, as \(|x^{-1}| = |x| > 2\) (Since order of a group element equals the order of its inverse.)

Hence there must be at least one element which has not been paired off with any of the others which is therefore self-inverse. Let’s say it , \('y'\).

\[\begin{eqnarray} y^{-1} =y &\iff & x\cdot y=e\\ &\iff & |y| =2 \end{eqnarray}\]

Thus, every group of even order contains an element of order \(2\).

Claim b: Let \(G\) be a group whose identity is \(e\).Let \(G\) be of order \(21\).(|G=21|) Then,\(\exists x \in G: |x| = 3\)

(Double Cosets) Let \(H\) and \(K\) be subgroups of a group \(G\), and let \(g\) be an element of \(G\). \ The set \(HgK = \{x \in G | x = hgk\) for some \(h \in H, k \in K\}\) is called a double coset. \ Do the double cosets partition \(G\)?

They partition G. The proof is similar to the proof that left cosets of one subgroup partition G.

On G, define the relation \(\sim\) as \[ a \sim b \text{ iff } b = hak; \text{ for some } h \in H, k \in K\] Let’s prove that this is an equivalence relation. Let \(a,b,c\in G\). - Reflexive?: Let \(g \in G\). Then we can write \[ g = 1g1\] Moreover, \(1 \in H\) and \(1 \in K\), so \(g \sim g\). Therefore, for every \(g \in G\) we have that \(g \sim g\). so \(\sim\) is reflexive.

• Symmetric?: Suppose that \(a \sim b.\) Then \(b = hak,\) for some \(h \in H, k \in K.\) Multiplying by\(h^{-1}\) from the left and\(k^{-1}\) from the right we get \[ a = h^{-1}ak^{-1}\] Since\(H\) is a subgroup of\(G,h^{-1} \in H.\) Similarly,\(k^{-1} \in K.\) Thus,\(b \sim a.\) So, for every\(a,b \in G\) such that \(a \sim b\implies b \sim a,\). Thus, \(\sim\) is symmetric.

• Transitive?: Suppose that \(a \sim b\) and \(b \sim c\).Then\(b = hak\)for some\(h \in H,k \in K\),and \(c=h'b'k'\) for some \(h' \in H,k' \in K\).Therefore, \[c=h'hbk'=hhakkk'= (hh')a(kk')\] Thus, \(a\sim b\) and \(b\sim c\) then \(a\sim c\). Therefore \(\sim\) is transitive.

By Artin’s book Proposition 2.7.4, double cosets partition G?

Let \(H\) be a subgroup of a group \(G\). Show that the double cosets (see Exercise M.9)) \[HgH = \{h_1gh_2 | h_1, h_2 \in H\}\] are the left cosets \(gH\) if and only if \(H\) is normal.

• \((\implies)\): Suppose that \(gH=HgH\). Let \(g\in G\) and \(h\in H\). \(ghg^{-1}\in H?\).

Problem

Tempoarray files for exercises

Chapter 3

Fields

Prove that the numbers of the form \(a + b\sqrt(2)\),where a and b are rational numbers, form a subfield of \(\mathbb{C}\)

Find the inverse of \(5\) modulo \(p\), for \(p= 7, 11, 13,\) and \(17\).

Compute the product polynomial \((x^3 + 3x^2 + 3x + 1)(x^4 + 4x^3 + 6x^2 + 4x + 1)\) when the coefficients are regarded as elements of the field \(\mathbb{F}_7\). Explain your answer.

Consider the system of linear equations \[\begin{bmatrix}6 & -3\\ 2 & 6\end{bmatrix} \begin{bmatrix}x_1\\ x_2\end{bmatrix} =\begin{bmatrix}3\\ 1\end{bmatrix}\]

1. Solve the system in \(\mathbb{F}_p\) when \(p = 5, 11,\) and \(17\).
2. Determine the number of solutions when \(p = 7\).

Determine the primes \(p\) such that the \[A =\begin{bmatrix} 1 & 2 & 0 \\ 0 & 3 & -1 \\ -2 & 0 & 2 \end{bmatrix}\] is invertible, when its entries are considered to be in \(\mathbb{F}_p\).

Solve completely the systems of linear equations \(AX = 0\) and \(AX = B\), where \[ A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & -1 & -1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \]

1. in \(\mathbb{Q}\),
2. in \(\mathbb{F}_2\),
3. in \(\mathbb{F}_3\),
4. in \(\mathbb{F}_7\)

By finding primitive elements, verify that the multiplicative group \(\mathbb{F}_P^\times\);is cyclic for all primes \(p < 20\).

Let \(p\) be a prime integer. (a) Prove Fermat’s Theorem: For every integer \(a\), \(a^p \equiv a modulo p\). (b) Prove Wilson’s Theorem: \((p -1)!\equiv -1(modulo p)\).

Determine the orders of the matrices \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) and \(\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\) in the group \(GL_2(\mathbb{F}_7)\).

Interpreting matrix entries in the field \(\mathbb{F}_2\), prove that the four matrices \[\begin{align*} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \end{align*}\] form a field.

Hint: You can cut the work down by using the fact that various laws are known to hold for addition and multiplication of matrices.

First Observe followings

• Addition:

\(+\)	\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)

• Closure ?: By above table we can verify Closure.
• Identity ?: Identity element is \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\).
• Inverse ? : Inverse of each elements is itself.(We can verify it from the above table.)
• Associativity ?: Associativity property we can get from the matrix addition.
• Commutative ?: We can verify communicativeness of addition from the above table.

Thus, it makes abelian group under \(+\).

• Multiplication:

\(*\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)	\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)	\(\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\)

• Closure ?: By above table we can verify Closure.
• Identity ?: Identity element is \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\).
• Inverse ? : Inverse of each elements is itself.(We can verify it from the above table.)
• Associativity ?: Associativity property we can get from the matrix multiplication
• Commutative ?: We can verify communicativeness of addition from the above table.

Thus, the set of non-zero elements of given set is an abelian group.

Distributive property: Multiplications is distributive over addition.

Thus, The given set forms a field.

Prove that the set of symbols \(\{a + bi \mid a, b \in \mathbb{F}_3\}\) forms a field with nine elements, if the laws of composition are made to mimic addition and multiplication of complex numbers. Will the same method work for \(\mathbb{F}_5\)? For \(\mathbb{F}_7\)? Explain.

VectorSpaces

2. 1. Prove that the scalar product of a vector with the zero element of the field \(F\) is the zero vector.

2. Prove that if \(w\) is an element of a subspace \(W\), then \(-w\) is in \(W\) too.

Which of the following subsets is a sub space of the vector space \(F^{n\times n}\) of \(n\times n\) matrices with coefficients in \(F\)?

1. symmetric matrices(\(A = A^T)\),
2. invertible matrices,
3. upper triangular matrices.

Find a basis for the space of \(n \times n\) symmetric matrices \((A^T = A)\).

Let \(W \subset \mathbb{R}^4\) be the space of solutions of the system of linear equations \(AX = 0\), where \(A =\begin{bmatrix} 2 & 1 & 2 & 3 \\ 1 & 1 & 3 & 0 \end{bmatrix}\). Find a basis for \(W\).

Prove that the three functions \(x^2, \cos(x)\), and \(e^x\) are linearly independent.

Suppose that \[\alpha x^2+ \beta \cos(x) +\gamma e^x =0\quad \text{for all } x\in \mathbb{R}\] where \(\alpha,\beta,\gamma\) scalers.

1. The original equation, when substitute \(x=0\) gives us \(b \cdot \cos(0) + c \cdot e^0 = 0\), which simplifies to \(b + c = 0\).
2. Differentiating once gives \(2a \cdot x - b \cdot \sin(x) + c \cdot e^x = 0\). Substituting \(x = 0\) gives \(-b + c = 0\).
3. Differentiating twice gives \(2a - b \cdot \cos(x) + c \cdot e^x = 0\). Substituting \(x = 0\) gives \(2a + b + c = 0\).

These equations show that the only solution is \(a = b = c = 0\). Therefore, the functions \(x^2\), \(\cos(x)\), and \(e^x\) are linearly independent.

Let \(A\) be an \(m \times n\) matrix, and let \(A'\) be the result of a sequence of elementary row operations on \(A\). Prove that the rows of \(A\) span the same space as the rows of \(A'\).

Vector spaces.